Induction motor rotor circuit model and energy conservation

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

• 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)

• 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)

• 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

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

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

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

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 %

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]}

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

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

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

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

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

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

• (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 ◦

• 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

• (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

• (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

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

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

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

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

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

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

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)

CHARACTERISTIC
• Development of induction motor torque –speed

Development of induction motor
torque –speed

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

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.

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?

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

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)]

TORQUE EQUATION
• Thevenin equivalent voltage of induction motor

Induction motor rotor circuit model and energy conservation

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