Uploaded bydatamboli
PPT, PDF170 views

Induction motor rotor circuit model and energy conservation

The document describes the torque-speed characteristic of induction motors. It explains that as the load on the motor increases, the slip increases, causing the rotor speed to decrease. This higher slip produces a stronger rotor voltage and current, increasing the rotor magnetic field BR. While BR increases torque, the increased angle δ between BR and the stator field Bnet acts to decrease torque. However, the increase in BR dominates, causing the overall induced torque to rise with increasing load and slip. The document derives an equation for induced torque as a function of slip to represent the torque-speed characteristic of induction motors.

Embed presentation

Download to read offline
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 speed as 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 
ROTOR CIRCUIT MODEL • Rotor voltage is ER0 constant & rotor impedance ZR,eq contains effects of varying slip
ROTOR CIRCUIT MODEL • Plot of current flow in rotor from equations: (1) & (2) shown below:
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 value as 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
FINAL EQUIVALENT CIRCUIT • Reminding: that in an ordinary transformer, voltages, currents and impedances on secondary side of device can be referred to primary side by 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)
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)
FINAL EQUIVALENT CIRCUIT • Rotor resistance RR & locked-rotor reactance XR0 are difficult or impossible to determine directly on cage rotors & effective turns ratio aeff also difficult to determine for cage rotors • fortunately, can make measurements that directly provide referred resistance and reactance R2 & X2 , (though RR, XR0 and aeff not known separately) • measurement of these parameters covered later
POWER & TORQUE IN INDUCTION MOTORS-LOSSES • since induction motor is 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
POWER & TORQUE IN INDUCTION MOTORS-LOSSES • Part of power transferred to rotor lost as : I^2 R=PRCL & rest converted from electrical to mechanical form Pconv, friction & windage losses PF&W & stray losses Pmisc subtracted  Pout
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
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 called rotational losses • Since component losses of rotational losses change 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
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 %
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]}
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 I2^2 R2 (2) • Pconv=PAG- PRCL=3I2^2 R2/s-3I2^2 R2 = 3I2^2 R2 (1/s-1) Pconv= 3I2^2 R2 (1-s)/s
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
POWER & TORQUE IN INDUCTION MOTORS • Induced torque Tind as : torque generated by internal electric to mechanical power conversion • It differs from available torque by amount equal to friction & windage torques in machine • Tind=Pconv/ωm also called developed torque of machine • Substituting for Pconv from (3) & for ωm, (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
POWER & TORQUE IN INDUCTION MOTORS • SEPARATION of PRCL & Pconv in induction motor Eq. cct. • Part of power coming across air gap consumed 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
POWER & TORQUE IN INDUCTION MOTORS • The per-phase equivalent circuit with rotor copper losses & power converted to mech. form separated into distinct elements shown below:
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: (a) Speed (b) stator current (c) P.F. (d) Pconv & Pout (e) Tind & Tload (f) Efficiency
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 ◦
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
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 W and the power converted is: • Pconv=(1-s)PAG =(1-0.022)(11845) =11585 W • Pout=Pconv-Prot=11585-1100=10485 W=14.1 hp
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 %
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 examined first from physical viewpoint of motor’s magnetic field & then a general equation for torque as function of slip derived
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
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
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
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 motion produces a stronger rotor voltage ER which in turn produces a larger rotor current IR
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 increased with load
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
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 torque is proportional to 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)
INDUCTION MOTOR TORQUE CHARACTERISTIC • Development of induction motor torque –speed
Development of induction motor torque –speed
Development of induction motor torque –speed
Development of induction motor torque –speed
INDUCTION MOTOR TORQUE CHARACTERISTIC • This characteristic curve can be divided into three regions • 1st region: is low-slip region in which motor slip increases approximately linearly with increase load & rotor mechanical speed decreases approximately linearly with load • In this region rotor 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
INDUCTION MOTOR TORQUE CHARACTERISTIC • 2nd region on curve called moderate-slip region • In moderate-slip region rotor frequency is 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 before and 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.
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 completely overshadowed 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
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?
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
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 E1 in 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)]
INDUCTION MOTOR INDUCED- TORQUE EQUATION • Thevenin equivalent voltage of induction motor

More Related Content

PDF
Four Point Starter: Diagram and Working Principle
byDr.Raja R
 
PPTX
Permanent magnet brushless dc motors ppt
bysrmrithi
 
PPTX
Deep bar n double cage rotors
byManish Sahu
 
PPTX
Brushless dc motor
byKUMARESAN ARUNAGIRI
 
PPT
Power driver circuits of stepper motor
bykarthi1017
 
PPTX
HYSTERESIS MOTOR
byragusundar
 
PDF
srm
byBadal Patnaik
 
PPTX
Stepper motor control
byJatin Arora
 
Four Point Starter: Diagram and Working Principle
byDr.Raja R
 
Permanent magnet brushless dc motors ppt
bysrmrithi
 
Deep bar n double cage rotors
byManish Sahu
 
Brushless dc motor
byKUMARESAN ARUNAGIRI
 
Power driver circuits of stepper motor
bykarthi1017
 
HYSTERESIS MOTOR
byragusundar
 
srm
byBadal Patnaik
 
Stepper motor control
byJatin Arora
 

What's hot

PPTX
Alternator and 3 phase induction motor
byjohny renoald
 
PPTX
Bldc motors ppt
byClint Robby
 
PDF
synchronous motor, Starting Torque, Types, Equivalent Circuit, Torque-speed c...
byWaqas Afzal
 
PPTX
Dc motors and its types
bySumeet Patel
 
PPT
Induction motors
byAbdulAziz Ahmed Siyad
 
PPTX
Generator and Transformer Protection (PART 1)
byDr. Rohit Babu
 
PPTX
Dc motor ppt
bysubhajit798bose
 
PPTX
brush less dc motor
byHarish Battula
 
PPTX
POWER QUALITY IMPROVEMENT BY DSTATCOM
bySajid Sheikh
 
PPTX
Electrical analogous of mechanical translational system
byKALPANA K
 
PPTX
three phase induction motor drive
byGuru Moorthi
 
PPTX
Stepper motor
byAnkit Basera
 
PPT
Dual converters (1)
byKarthiKeyani10
 
PPT
instrument transformer
byNishant Kumar
 
PDF
Induction motors
bygudlavalleru engineering college
 
PDF
Closed Loop Control of DC Drives
byRohanSharma344025
 
PPTX
3. INTRODUCTION TO PROTECTIVE RELAYING.pptx
byMuhd Hafizi Idris
 
PPT
Induction motor three phase
byStudy
 
PPTX
Selection criteria of motors for various applications
byPrasanth Ravula
 
PPTX
Definition & Classification Of Power System Stability
byShahab Khan
 
Alternator and 3 phase induction motor
byjohny renoald
 
Bldc motors ppt
byClint Robby
 
synchronous motor, Starting Torque, Types, Equivalent Circuit, Torque-speed c...
byWaqas Afzal
 
Dc motors and its types
bySumeet Patel
 
Induction motors
byAbdulAziz Ahmed Siyad
 
Generator and Transformer Protection (PART 1)
byDr. Rohit Babu
 
Dc motor ppt
bysubhajit798bose
 
brush less dc motor
byHarish Battula
 
POWER QUALITY IMPROVEMENT BY DSTATCOM
bySajid Sheikh
 
Electrical analogous of mechanical translational system
byKALPANA K
 
three phase induction motor drive
byGuru Moorthi
 
Stepper motor
byAnkit Basera
 
Dual converters (1)
byKarthiKeyani10
 
instrument transformer
byNishant Kumar
 
Induction motors
bygudlavalleru engineering college
 
Closed Loop Control of DC Drives
byRohanSharma344025
 
3. INTRODUCTION TO PROTECTIVE RELAYING.pptx
byMuhd Hafizi Idris
 
Induction motor three phase
byStudy
 
Selection criteria of motors for various applications
byPrasanth Ravula
 
Definition & Classification Of Power System Stability
byShahab Khan
 

Similar to Induction motor rotor circuit model and energy conservation

PDF
Induction motors
byUthsoNandy
 
PPTX
Induction machines
byMohammed Waris Senan
 
PPT
Induction Machines.ppt
byBaladuraikannan thiyagarajan
 
PPT
Induction Machine electrical and electronics
byprakashpacet
 
PPT
Rotating Electrical Machines-AC & DC Machines,Induction Motor and DC Motor
byPrasant Kumar
 
PPT
Induction machines
by1234marvin
 
PPT
Induction machines
by1234marvin
 
PDF
EM_II_PPT.pdf
byDailyMeditation
 
PPT
Induction Machines electrical machines electrical and electronics
byprakashpacet
 
PPTX
Electrical Power Systems Induction motor
byMubarek Kurt
 
PPT
EE-260 Lecture 27, 28. The equivalent circuit of an induction motor
byMuneebTahir24
 
PDF
three phase induction motor and starting methods
byssuserc6442c
 
DOCX
Induction machine
bySURAJ PRASAD
 
PPT
Induction_Machines.ppt all details concept
byManishKhaniya2
 
PPT
Introduction to Electrical Machines, Induction Machines
byhilinan21
 
PPT
induction_motor1_1_1_8space vector _ppt.ppt
byBrajeshSingh954924
 
PPTX
Determination of Equivalent Circuit parameters and performance characteristic...
bypvpriya2
 
PPT
Induction and synchronous Motor drives (AC motor drives)
byOubella M'hand
 
PPTX
Lecture_9_Three_Phase_Induction_Machine_Intro.pptx
byntofficekalikot
 
PPT
Induction machines
byTANUSHREE BANIK
 
Induction motors
byUthsoNandy
 
Induction machines
byMohammed Waris Senan
 
Induction Machines.ppt
byBaladuraikannan thiyagarajan
 
Induction Machine electrical and electronics
byprakashpacet
 
Rotating Electrical Machines-AC & DC Machines,Induction Motor and DC Motor
byPrasant Kumar
 
Induction machines
by1234marvin
 
Induction machines
by1234marvin
 
EM_II_PPT.pdf
byDailyMeditation
 
Induction Machines electrical machines electrical and electronics
byprakashpacet
 
Electrical Power Systems Induction motor
byMubarek Kurt
 
EE-260 Lecture 27, 28. The equivalent circuit of an induction motor
byMuneebTahir24
 
three phase induction motor and starting methods
byssuserc6442c
 
Induction machine
bySURAJ PRASAD
 
Induction_Machines.ppt all details concept
byManishKhaniya2
 
Introduction to Electrical Machines, Induction Machines
byhilinan21
 
induction_motor1_1_1_8space vector _ppt.ppt
byBrajeshSingh954924
 
Determination of Equivalent Circuit parameters and performance characteristic...
bypvpriya2
 
Induction and synchronous Motor drives (AC motor drives)
byOubella M'hand
 
Lecture_9_Three_Phase_Induction_Machine_Intro.pptx
byntofficekalikot
 
Induction machines
byTANUSHREE BANIK
 

More from datamboli

PPT
Single phase induction motor-Working principle types.ppt
bydatamboli
 
PPTX
single phase Induction Motor-types, construction working.pptx
bydatamboli
 
PPT
Starting and speed control of induction motor.ppt
bydatamboli
 
PPTX
Single-phase-Induction-motor1.ppt.pptx
bydatamboli
 
PPTX
DC machine, construction and armature reaction.pptx
bydatamboli
 
PPT
Introduction to AC Motors with constructional details.ppt
bydatamboli
 
PPTX
Unit 1-Introduction of Control system.pptx
bydatamboli
 
PPTX
Unit 3-State-Space Analysis_all.pptx
bydatamboli
 
PPT
single phase induction motor working and types.ppt
bydatamboli
 
PPTX
Power Factor Correction in Industry for Improving Efficiency.pptx
bydatamboli
 
PPTX
lec-7_phase_plane_analysis.pptx
bydatamboli
 
PPTX
L2 unit 1-definition and practical examples, basic clcs
bydatamboli
 
PPT
ENERGY_CONVERSION)During Motor Principle.ppt
bydatamboli
 
Single phase induction motor-Working principle types.ppt
bydatamboli
 
single phase Induction Motor-types, construction working.pptx
bydatamboli
 
Starting and speed control of induction motor.ppt
bydatamboli
 
Single-phase-Induction-motor1.ppt.pptx
bydatamboli
 
DC machine, construction and armature reaction.pptx
bydatamboli
 
Introduction to AC Motors with constructional details.ppt
bydatamboli
 
Unit 1-Introduction of Control system.pptx
bydatamboli
 
Unit 3-State-Space Analysis_all.pptx
bydatamboli
 
single phase induction motor working and types.ppt
bydatamboli
 
Power Factor Correction in Industry for Improving Efficiency.pptx
bydatamboli
 
lec-7_phase_plane_analysis.pptx
bydatamboli
 
L2 unit 1-definition and practical examples, basic clcs
bydatamboli
 
ENERGY_CONVERSION)During Motor Principle.ppt
bydatamboli
 

Recently uploaded

PPTX
Unit 5 - THERMO ELECTRIC ENERGY BASED PROCESSES .pptx
byarivazhaganrajangam
 
PPTX
Introduction to Control Systems Chapter 2.pptx
byJosephHassen
 
PPTX
TR334_Foundation_Engineering_I_Slope stability _i.pptx
byalexander402774
 
PPT
Steam generation , types of steam, boilers and their classifications, mountin...
byPrashantPatunkar1
 
PPTX
Basics of internal combustio nengines, vapor compression cycle
byPrashantPatunkar1
 
PPTX
narrow-band-dfr-presentation-megger.pptx
byssuser52ed331
 
PDF
RAMON Scrap quality detection system.pdf
bymohamedassar81
 
PPTX
Forouzan6e_ch02_PPTs_Physical.pptxmcgrawhill
byshaiviadesh
 
PPT
Unit 1 - SHAPER, MILLING AND GEAR CUTTING MACHINES .ppt
byarivazhaganrajangam
 
PPTX
chapterOverview - embedded systems .pptx
bysamyukthakg2020
 
PDF
Bme manufacturing process module 3 .4648
bysatyabsp44
 
PDF
Module 2 Python progr 1BPLCK105B-by Dr.SV.pdf
bySURESHA V
 
PPTX
SOFTWARE PROJECT MANAGEMENT PART 2-2.pptx
byssuserbafe2d
 
PPT
CHAPTER 09 - Digital signatures cryptography note
bySherin Wilson
 
PDF
Unconventional reservoirs in petroleum engineering
byMuhammadHaris61480
 
PPTX
UNIT-1.pptx...................................
byRavinderKaur647214
 
PPTX
SOFTWARE PROJECT MANAGEMENT PART 1-.pptx
byssuserbafe2d
 
PPTX
STM Unit 2 Complete notes for engineering students for better understanding
byLakshmiVivekaKesanap
 
PPTX
Lecture 2_Introduction to engineering design.pptx
bylynxvill3
 
PPT
Introduction to ARTIFICIAL INTELLIGENCES
byKAVITHAA49
 
Unit 5 - THERMO ELECTRIC ENERGY BASED PROCESSES .pptx
byarivazhaganrajangam
 
Introduction to Control Systems Chapter 2.pptx
byJosephHassen
 
TR334_Foundation_Engineering_I_Slope stability _i.pptx
byalexander402774
 
Steam generation , types of steam, boilers and their classifications, mountin...
byPrashantPatunkar1
 
Basics of internal combustio nengines, vapor compression cycle
byPrashantPatunkar1
 
narrow-band-dfr-presentation-megger.pptx
byssuser52ed331
 
RAMON Scrap quality detection system.pdf
bymohamedassar81
 
Forouzan6e_ch02_PPTs_Physical.pptxmcgrawhill
byshaiviadesh
 
Unit 1 - SHAPER, MILLING AND GEAR CUTTING MACHINES .ppt
byarivazhaganrajangam
 
chapterOverview - embedded systems .pptx
bysamyukthakg2020
 
Bme manufacturing process module 3 .4648
bysatyabsp44
 
Module 2 Python progr 1BPLCK105B-by Dr.SV.pdf
bySURESHA V
 
SOFTWARE PROJECT MANAGEMENT PART 2-2.pptx
byssuserbafe2d
 
CHAPTER 09 - Digital signatures cryptography note
bySherin Wilson
 
Unconventional reservoirs in petroleum engineering
byMuhammadHaris61480
 
UNIT-1.pptx...................................
byRavinderKaur647214
 
SOFTWARE PROJECT MANAGEMENT PART 1-.pptx
byssuserbafe2d
 
STM Unit 2 Complete notes for engineering students for better understanding
byLakshmiVivekaKesanap
 
Lecture 2_Introduction to engineering design.pptx
bylynxvill3
 
Introduction to ARTIFICIAL INTELLIGENCES
byKAVITHAA49
 

Induction motor rotor circuit model and energy conservation

  • 1.
    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 speed as 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 
  • 2.
    ROTOR CIRCUIT MODEL •Rotor voltage is ER0 constant & rotor impedance ZR,eq contains effects of varying slip
  • 3.
    ROTOR CIRCUIT MODEL •Plot of current flow in rotor from equations: (1) & (2) shown below:
  • 4.
    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 value as 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
  • 5.
    FINAL EQUIVALENT CIRCUIT •Reminding: that in an ordinary transformer, voltages, currents and impedances on secondary side of device can be referred to primary side by 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)
  • 6.
    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)
  • 7.
    FINAL EQUIVALENT CIRCUIT •Rotor resistance RR & locked-rotor reactance XR0 are difficult or impossible to determine directly on cage rotors & effective turns ratio aeff also difficult to determine for cage rotors • fortunately, can make measurements that directly provide referred resistance and reactance R2 & X2 , (though RR, XR0 and aeff not known separately) • measurement of these parameters covered later
  • 8.
    POWER & TORQUEIN INDUCTION MOTORS-LOSSES • since induction motor is 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
  • 9.
    POWER & TORQUEIN INDUCTION MOTORS-LOSSES • Part of power transferred to rotor lost as : I^2 R=PRCL & rest converted from electrical to mechanical form Pconv, friction & windage losses PF&W & stray losses Pmisc subtracted  Pout
  • 10.
    POWER & TORQUEIN 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
  • 11.
    POWER & TORQUEIN 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 called rotational losses • Since component losses of rotational losses change 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
  • 12.
    POWER & TORQUEIN 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 %
  • 13.
    POWER & TORQUEIN 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]}
  • 14.
    POWER & TORQUEIN 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 I2^2 R2 (2) • Pconv=PAG- PRCL=3I2^2 R2/s-3I2^2 R2 = 3I2^2 R2 (1/s-1) Pconv= 3I2^2 R2 (1-s)/s
  • 15.
    POWER & TORQUEIN 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
  • 16.
    POWER & TORQUEIN INDUCTION MOTORS • Induced torque Tind as : torque generated by internal electric to mechanical power conversion • It differs from available torque by amount equal to friction & windage torques in machine • Tind=Pconv/ωm also called developed torque of machine • Substituting for Pconv from (3) & for ωm, (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
  • 17.
    POWER & TORQUEIN INDUCTION MOTORS • SEPARATION of PRCL & Pconv in induction motor Eq. cct. • Part of power coming across air gap consumed 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
  • 18.
    POWER & TORQUEIN INDUCTION MOTORS • The per-phase equivalent circuit with rotor copper losses & power converted to mech. form separated into distinct elements shown below:
  • 19.
    POWER & TORQUEIN 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: (a) Speed (b) stator current (c) P.F. (d) Pconv & Pout (e) Tind & Tload (f) Efficiency
  • 20.
    POWER & TORQUEIN 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 ◦
  • 21.
    POWER & TORQUEIN 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
  • 22.
    POWER & TORQUEIN 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 W and the power converted is: • Pconv=(1-s)PAG =(1-0.022)(11845) =11585 W • Pout=Pconv-Prot=11585-1100=10485 W=14.1 hp
  • 23.
    POWER & TORQUEIN 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 %
  • 24.
    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 examined first from physical viewpoint of motor’s magnetic field & then a general equation for torque as function of slip derived
  • 25.
    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
  • 26.
    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
  • 27.
    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
  • 28.
    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 motion produces a stronger rotor voltage ER which in turn produces a larger rotor current IR
  • 29.
    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 increased with load
  • 30.
    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
  • 31.
    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 torque is proportional to 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)
  • 32.
    INDUCTION MOTOR TORQUE CHARACTERISTIC •Development of induction motor torque –speed
  • 33.
    Development of inductionmotor torque –speed
  • 34.
    Development of inductionmotor torque –speed
  • 35.
    Development of inductionmotor torque –speed
  • 36.
    INDUCTION MOTOR TORQUE CHARACTERISTIC •This characteristic curve can be divided into three regions • 1st region: is low-slip region in which motor slip increases approximately linearly with increase load & rotor mechanical speed decreases approximately linearly with load • In this region rotor 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
  • 37.
    INDUCTION MOTOR TORQUE CHARACTERISTIC •2nd region on curve called moderate-slip region • In moderate-slip region rotor frequency is 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 before and 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.
  • 38.
    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 completely overshadowed 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
  • 39.
    INDUCTION MOTOR INDUCED- TORQUEEQUATION • 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?
  • 40.
    INDUCTION MOTOR INDUCED- TORQUEEQUATION • 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
  • 41.
    INDUCTION MOTOR INDUCED- TORQUEEQUATION • 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 E1 in 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)]
  • 42.
    INDUCTION MOTOR INDUCED- TORQUEEQUATION • Thevenin equivalent voltage of induction motor