ENERGY CONVERSION ONE
(Course 25741)
CHAPTER SEVEN
…..INDUCTION MOTORS
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 INDUCEDTORQUE 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 INDUCEDTORQUE 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 INDUCEDTORQUE 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 INDUCEDTORQUE EQUATION
• Thevenin equivalent voltage of induction motor