presentation on power and torque of induction motor

Presentation On Power & Torque in Induction Motors

Presented ByWajahat Rafique Chohan

Nouman Ahmed

Hasnain Haroon

Hafiz Muhammad Umer

Syed Wajahat Ali

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


• 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


• 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


• 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

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]}


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


• 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


• 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


• 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


• The per-phase equivalent circuit with rotor copper losses & power converted to mech. form separated into distinct elements shown below:

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


• Induced Torque from a Physical Viewpoint

• Figure shows a cage rotor of an induction motor

• Initially operating at no load & nearly sync. speed


• 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


• 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


• 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


• 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


• 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


• 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)


• Development of induction motor torque –speed

Development of induction motor torque –speed


• 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


• 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.


• 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?


• 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


• 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)]


• Thevenin equivalent voltage of induction motor

presentation on power and torque of induction motor

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