Outline
1
Electrical Drive Systems 324
Induction Machines
Dr. P.J Randewijk
Stellenbosch University
Dep. of Electrical & Electronic Engineering
Stephan J. Chapman
Chapter 6 (5th Edition)
Chapman, Chapter 6
6.1 – Induction Motor Construction
6.2 – Basic Induction Machine Concepts
6.3 – The Equivalent Circuit of an Induction Motor
6.4 – Power and Torque in Induction Motors
6.5 – Induction Motor Torque–Speed Characteristics
6.6 – Variations in Induction Motor Torque–Speed
Characteristics
6.8 – Starting Induction Motors
6.9 – Speed Control of Induction Motors
6.10 – Solid-State Induction Motor Drives (or VSDs)
6.11 – Determining (of the Equivalent) Circuit Model
Parameters
6.12 – The Induction Generator
6.13 – Induction Motor Ratings
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6.1 Induction Machine Construction
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6.2 Basic Induction Machine Concepts
The stator of an induction (sometimes also called an
asynchronous) machine is exactly the same as that of a
synchronous machine
The only difference is in the rotor design
Two types of rotor designs exist – must know the
difference between the two designs:
The Development of Induced Torque
The magnetic field produced by the three-phase stator
winding, when a three-phase current is supplied to them,
BS , rotates at synchronous speed – see also eq. (3–34)
nsync = 60fse ·
cage rotor
wound rotor
2
poles
(6–1’)
At standstill, the rotating BS will induce a voltage in the
stationary rotor windings/bars
It is important to note that the rotor windings are all
short-circuited!
eind = (v × B) · l
(1–45)
Because the rotor windings/bars are short-circuited, a
current will start to flow in the rotor windings/bars
The current, in the presence of BS will produce a force
on the windings/bars in the rotor in the direction of the
rotating BS
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6.2 Basic Induction Machine Concepts (cont.)
The induced force will start to accelerate the rotor in the
direction of the rotating BS
+ Note: The rotor will never be able to rotate at
synchronous speed, because then the induced voltage,
and hence the current flowing in the rotor and hence the
induced force acting on the rotor windings/bars will be
zero. . .
The Concept of Rotor Slip
The term more often used is slip – sometimes
expressed as a percentage – defined as
s=
s=
nslip
(×100 %)
nsync
nsync − nm
(×100 %)
nsync
(6–3)
(6–4)
Slip can also be calculated form the speed in [rad/s]
Because the rotor of an induction motor is rotating at a
slower speed than that of the synchronous speed of the
rototing BS
We say that the rotor is slipping
The slip speed is the difference in the synchronous
speed and the rotor speed
nslip = nsync − nm
6.2 Basic Induction Machine Concepts (cont.)
(6–2)
The Electrical Frequency on the Rotor
An induction machine can be viewed as a rotating
transformer
At standstill, s = 1, and the frequency of the induced
rotor voltage and current will be the same as that of the
stator frequency
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6.2 Basic Induction Machine Concepts (cont.)
If the rotor were to rotate at synchronous speed, i.e.
s = 0, the frequency of the induced rotor voltage will
0 Hz. . .
The rotor frequency can be generalised as:
fre = sfse
poles nslip
·
2
60
6.3 The Equivalent Circuit
The Transformer Model of an Induction Machine
(6–8)
Or from eq. (6–1’) and (6–2)
fre =
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An induction machine is singly excited vs. the doubly
excited synchronous and DC machines
There will therefore be no internal generated voltage,
EA , as for a synchronous machine
The equivalent circuit looks exactly the same as that of a
transformer – see Fig. 6–07
(6–9’)
+ Note: The rotating magnetic field produced by the rotor,
BR , will always rotate at the same synchronous speed
as the magnetic field of the stator, BS .
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6.3 The Equivalent Circuit (cont.)
6.3 The Equivalent Circuit (cont.)
From the stator side, the only difference with that of a
transformer model would be that the value of XM would
be much smaller due to the airgap between the stator
and rotor
The B–H curve would also be much flatter than for a
transformer, i.e. require more current to generate the
same flux-density in the machine – see Fig. 6–8
This implies that the rotor circuit parameters can be
expressed (see proof) in terms of the stator frequency
(e.g. 50 Hz) values
The Final Equivalent Circuit
We can now refer the rotor circuit to the stator circuit,
taking into account the (effective) turns ratio,
The Rotor Circuit Model
aeff =
The induced voltage in the rotor, ER , will be s× the
locked/blocked-rotor voltage, ER0
ER = sER0
so that we can write
(6–10)
Also from eq. (6–8), the reactance in the rotor circuit will
vary with frequency, so that
XR = sXR0
Ns
Nr
(6–11)
E1 = E0R = aeff ER0
IR
I2 =
aeff
R
R
2
Z2 = aeff
+ jXR0
s
(6–18)
(6–19)
(6–20)
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6.3 The Equivalent Circuit (cont.)
6.4 Power and Torque
And by making the following definitions
R2 =
X2 =
2
aeff
RR
2
aeff XR0
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Losses and the Power-Flow Diagram
(6–21)
(6–22)
the per-phase equivalent circuit of the induction machine
as referred to the stator
Power and Torque in an Induction Machine
The stator copper losses can be calculated as follows
PSCL = 3I12 R1
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(6–25)
12 / 53
6.4 Power and Torque (cont.)
6.4 Power and Torque (cont.)
The core losses (if RC is given and the core losses is not
lumped together with the friction-and-windage losses
Pcore =
3E12
RC
PAG = Pin − PSCL − Pcore
PAG =
R2
3I22
s
PRCL = 3I22 R2
(6–26’)
The power transferred from the stator over the air-gap to
the rotor
which is identical to the power transferred to
With R2 the value of the rotor circuit resistance referred
to the stator, the rotor copper losses will thus be equal to
The power converted to mechanical power, the
developed mechanical power will thus be
Pconv = PAG − PRCL
1 − s
Pconv = 3I22 R2
s
(6–27)
R2
s
(6–28)
Pconv
ωm
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Pout = Pconv − PF&W − Pmisc
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Seperating PRCL and Pconv in the Equivalent Circuit
(6–34)
This implies that the output torque will be
Pconv = (1 − s)PAG
(6–33)
And with ωm = (1 − s)ωsync , the induced torque can be
expressed as a function of the air-gap power as
PAG
ωsync
The resistor, Rs2 is sometimes “split” or separated into
two separate resistors, R2 and Rconv
This is done in order to illustrate the fact that the airgap
power, PAG associated with Rs2 can also be separated
into
PRCL associated with R2 and
Pconv associated with Rconv
Pout
ωm
Furthermore, by noting that PRCL = sPAG , it follows that
τind =
(6–35)
6.4 Power and Torque (cont.)
+ Note: The output torque, say τout will be less than the
developed torque due to the output power,
τout =
(6–31)
The induced torque or developed torque can now be
calculated as
τind =
6.4 Power and Torque (cont.)
(6–30)
Thus with
∵
∴
Pconv = PAG − PRCL
R2
Rconv =
− R2
s
1−s
= R2
s
(6–37)
(6–36)
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6.4 Power and Torque (cont.)
6.5 Torque–Speed Characteristics
+ N.B. The crib sheet for all induction machine
calculations can thus be summarised by the following
equivalent circuit diagram with the associated powers
Induced Torque from a Physical Standpoint – read
only
The Derivation of the Induction Machine
Induced-Torque Equation ⇔ Solving the Equivalent
Circuit of the Induction Machine
The core –, Pcore , the friction-and-windage –, PF&W and
the miscellaneous or stray losses, Pstray , are usually
lumped together and called the rotational losses, Prot
This is done, due to the fact that is it very difficult to
separate them during the determination of the equivalent
parameters of the induction machine – see section 6.11
This implies that the equivalent circuit of the induction
machine can be simplified to that shown in Fig. 6–17,
with the equivalent core loss resistance, RC , ignored. . .
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6.5 Torque–Speed Characteristics (cont.)
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6.5 Torque–Speed Characteristics (cont.)
This implies that eq. (6-34) changes to
Pout = Pconv − Prot
(6–34’)
+ Note: See Example 6–3,
The easiest way to solve I2 in order to calculate PAG ,
Pconv and τind
is to firstly calculate the Thévenin equivalent between
the “X’s” in Fig. 6–17
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6.5 Torque–Speed Characteristics (cont.)
6.5 Torque–Speed Characteristics (cont.)
then I2 by noting
I2 =
VTH
RTH + jXTH + jX2 +
R2
s
(6–46’)
This allow us to calculate PAG from eq. (6–30)
Pconv from eq. (6–33) and
τind from eq. (6–36)
The maximum torque that the machine can develop is
called the pullout torque or breakdown torque
The starting torque is only slightly higher than the
full-load torque, compared to the starting torque of a DC
machine which can be extremely high. . .
+ See Fig. 6–19 (next slide)
If the rotor is turning faster than the synchronous speed,
the induction machine operates as a generator
If the rotor is turning in the opposite direction as that of
the rotating field on the stator, the motor will be braked –
this is called plugging
+ See Fig. 6–20 (subsequent slide)
+ Note: Do not try and remember or use eq. (6–41b),
(6–44), (6–45), (6–49) or (6–50) !!!
Comments on the Induction Machine’s
Torque-Speed Curve
At synchronous speed the induced torque is zero
The torque–speed curve is near linear between no-load
and full-load
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6.5 Torque–Speed Characteristics (cont.)
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6.5 Torque–Speed Characteristics (cont.)
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6.5 Torque–Speed Characteristics (cont.)
6.5 Torque–Speed Characteristics (cont.)
Maximum (Pullout) Torque in an Induction Machine
With τind ∝ PAG , it implies that the maximum induced
torque will occur when the power consumed by R22 is a
maximum
The maximum power transfer theorem state that when
the magnitude of the source impedance is equal to the
magnitude of the load impedance
Thus with
Zsource = RTH + jXTH + jX2
(6–51)
This implies that the slip at pullout torque will be
R2
smax = q
2 + (X
2
RTH
TH + X2 )
R2
s
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6.5 Torque–Speed Characteristics (cont.)
ωsync
3V 2
q TH
2 + (X
2
RTH + RTH
TH + X2 )
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6.5 Torque–Speed Characteristics (cont.)
It can be proved from (6–50) that the maximum τind or
pullout torque can also be calculated with the following
“formula”
τind =
(6–53)
The maximum or pullout torque can now be calculated
by substituting the value of smax into eq. (6–28) to
calculate PAG
And then substituting PAG into eq. (6–36) to calculate
the maximum τind or pullout torque value
and
Zload =
Maximum power transfer (and hence maximum torque)
will occur when
q
R2
2 + (X
2
(6–52)
= RTH
TH + X2 )
s
(6–54)
+ Note: Please do not try and remember eq. (6–54)
However what is interesting to note from (6–54) is that
the maximum torque is independent of R2
Thus in a wound rotor induction machine, the value of R2
can effectively be manipulated by adding external
resistance via the slipring connections.
From (6–53) we can see that this will effectively change
the slip – and hence the speed – at which maximum
torque occurs.
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+ See Fig. 6–22 to see how the characteristic curve of the
machine can be varied by adding external resistance. . .
+ Also note that the characteristic curve of induction
machine with (external) rotor resistance control will look
identical to that of a DC machine with (external)
armature resistance control – see Fig. 8–15 with the y–
and x–axis’ reversed, i.e. τind vs. ωm ,
External rotor resistance control could thus also be used
to perform speed control on wound induction
machines. . . albeit not very efficient due to the
additional losses in the external rotor resistors. . .
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6.5 Torque–Speed Characteristics (cont.)
6.6 Variations in Motor Characteristics
By varying the design of the cage rotor, different
torque–speed characteristics can be obtained
. . . read only – also section 6.7 . . .
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6.8 Starting Induction Motors
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6.8 Starting Induction Motors (cont.)
Induction motors – unlike synchronous motors – can be
connected directly on line (DOL) or across-the-line to
start them
For cage rotor induction machines, the only way to
reduce the starting current, is to reduce the supply
voltage during start-up
Although the starting current and torque is not a high as
for DC motors, the starting current could be up to 5× the
rated current of the motor
Unfortunately this also leads to a reduced starting
torque, as the starting torque decrease with the square
of the applied voltage, as can be seen from (6–54)
Ignore the “Nominal code letters” of Fig. 6–34
The different “reduced voltage starting methods” are:
Wound rotor induction machines can be started by
adding extra resistance in the rotor circuit during starting
– this not only lowers the starting current, but also
increases the starting torque so that the load can be
quicker brought up to speed. . .
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inserting a resistor in series with the supply during
starting
inserting an inductor in series with the supply during
starting
using a autotransformer to reduce the starting voltage
using a soft starter – will be discussed under “Power
Electronics”
using
a Y–∆ starting circuit – the starting voltage is thus
√
3 lower during starting. . .
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6.8 Starting Induction Motors (cont.)
6.9 Speed Control of Induction Motors
+ DOL starting, plus the rest of this section will be
discussed in more detail under “Industrial Automation”
using IEC symbols instead of the ANSI sybols used by
Chapman.
Induction Motor Speed Control by Pole Changing –
read only
Speed Control by Changing the Line Frequency
Induction motor usually operate with a small slip value,
i.e. operate just below synchronous speed
By varying the line frequency, the synchronous speed
and hence operating speed can be varied
nsync = 60 · fse ·
2
poles
(6–1)
However by changing the frequency, the magnitude of
the magnetising inductance, XM , also changes
(X = 2πfL)
When we want to operate the motor at lower speed, this
would imply that the motor needs to be operated at a
lower frequency. . .
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6.9 Speed Control of Induction Motors (cont.)
A lower frequency would result in a lower value for XM ,
and hence a higher magnetising current. . .
A higher magnetising current would imply that the motor
will saturate. . .
To circumvent this, the supply voltage also needs to be
lowered with (roughly) the same ratio as that of the
frequency in order for the magnetising current to remain
the same
This is called – constant volts-per-hertz control. . .
All of this is usually done automatically using a Variable
Speed Drive or VSD
By keeping the magnetising current constant whilst
varying the frequency below its rated frequency,
constant torque operation from standstill to rated speed
can be achieved – similar to that of a DC machine, see
Fig. 8–16
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6.9 Speed Control of Induction Motors (cont.)
+ The general rule of thumb is that with the magnetising
current constant, the induced torque, τind , will be the
same if the slip speed, nsync −nm , is the same. . .
At rated speed, the supply voltage will be equal to its
rated value
Thus when the frequency is increase above its rated
frequency in order to operate the machine at above its
rated speed, it is usually impossible to increase the
supply voltage as well in order to keep the magnetising
current constant. . .
Hence the induced torque at speeds above rated speed
decrease hyperbolically (τind = Pωconv
) whilst the
m
developed power, Pconv , remains constant – similar to
that of a DC machine, see Fig. 8–16
+ Constant Volts-per-Hertz control is shown graphically in
Fig. 6–43 (a)–(c)
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6.9 Speed Control of Induction Motors (cont.)
6.9 Speed Control of Induction Motors (cont.)
The voltage drop across R1 at low frequencies leads to a
lower voltage across XM – hence a lower magnetising
current lower torque at low frequencies. . .
To compensate for this, VSDs employ “boost voltage”
control at lower frequencies – see Fig. 6–51
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6.9 Speed Control of Induction Motors (cont.)
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6.9 Speed Control of Induction Motors (cont.)
Speed Control by Changing the Line Voltage
The induced torque is proportion to the square of the
supply voltage, so this is not a very linear control method
– see Fig. 6–44
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6.9 Speed Control of Induction Motors (cont.)
Speed Control by Changing the Rotor Resistance
Not very efficient and you require an expensive wound
rotor machine – see Fig. 6–45
6.10 Solid-State Drives (or VSDs)
This was discussed in the previous section and will be
expanded during our discussion on Power Electronics
later in the course. . .
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6.11 Model Parameters (for Prac only)
6.11 Model Parameters (for Prac only) (cont.)
+ N.B. The procedure discussed in this section is a very
rough simplification
The DC Test for Stator Resistance
This is done to determine the stator resistance
By doing this test at the rated current value, the heating
effect on the resistance of the copper windings is taking
into account
For a Y–connected stator
R1 =
VDC
2IDC
(6–61)
For a ∆–connected stator (Why?)
R1 =
42 / 53
The same result could also be obtained by measuring
the stator resistance with a multimeter directly after the
locked-rotor test (so that the windings are nicely heated
up) and obviously again taking the connection of the
stator into account. . .
The Locked-Rotor Test
With s = 1 when the rotor is locked, the following
assumption (after years of experience) can be made
XM |R2 + jX2 |
RC |R2 + jX2 |
so that RC and XM can be neglected and a simple series
circuit remains
3 VDC
·
2 IDC
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6.11 Model Parameters (for Prac only) (cont.)
6.11 Model Parameters (for Prac only) (cont.)
and
XLR =
Qin =
(6–65)
if the test was conducted at a lower test frequency that
represents the normal operating condition of the rotor
more accurately
With Pin , VT and IL measured
√
Sin = 3VT IL
q
Qin = Sin2 − Pin2
Pin =
frated 0
X = X1 + X2
ftest LR
For our Practical however,
frated = ftest
3IL2 RLR
0
3IL2 XLR
∴
0
XLR
= XLR
this results in
RLR = R1 + R2
(6–65)
45 / 53
6.11 Model Parameters (for Prac only) (cont.)
The values for X1 and X2 (after years of experience) are
split as follows
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6.11 Model Parameters (for Prac only) (cont.)
The No-Load Test
With s ≈ 1 when the machine is running at no-load, the
following assumption can be made (after years of
experience)
RF&W XM
RC XM
so that RF&W and RC can be neglected and a simple
series circuit remains
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6.11 Model Parameters (for Prac only) (cont.)
With Pin , VT =
√
3Vφ and IL = I1 measured
√
Sin = 3VT IL
q
Qin = Sin2 − Pin2
Which implies that the machine’s rotational losses will be
equal to
Prot = Pin − PSCL
= Pcore + PF&W + Pmisc
= 3I12 XNL
(6–58’)
(6–59)
For our Practical however, PF&W is overcome by the DC
motor spinning the rotor. . .
Also if we were to ignore Pmisc , we can say that
This implies that
XNL = X1 + XM
Pcore = Pin − PSCL
+ N.B. This approach is slightly more accurate than eq.
(6–60) used in Chapman.
Furthermore, the Stator Copper Loss (SCL) can be
calculated as follows
PSCL = 3I12 R1
6.11 Model Parameters (for Prac only) (cont.)
=
(6–25)
3Vφ2
RC0
which can thus be modelled by an equivalent resistance,
RC0 , in parallel with the supply. . .
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6.12 The Induction Generator
50 / 53
6.12 The Induction Generator (cont.)
As soon as the rotor spins faster than synchronous
speed, power is absorbed from the mechanical side and
the machine operates as a generator
It is however important to note that the machine must be
connected to a three-phase supply in order to generate
a rotating MMF in the stator windings, for the machine to
operate as a generator.
Capacitors are sometimes used for stand alone / off-grid
applications, but is very “unstable”
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6.13 Induction Motor Ratings
Usually only the following information is given on the
nameplate
Rated (output) power, usually in multiples of 750 W. . .
Rated (line-to-line) voltage
Rated current
Rated frequency
Stator connection, whether Y or ∆
Rated speed
+ From the latter, the number of pole can be worked out
as the rated speed is usually just a little bit less than the
synchronous speed. . .
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