Transcript Microelectromechanical Devices

ECE 8830 - Electric Drives
Topic 3: Induction Motor Modeling Steady State
Spring 2004
Introduction
Induction machines are the most widely
used of all electric motors. They offer the
following attractive features:







Generally easy to build and cheaper than
corresponding dc or synchronous motors
Rugged and require little maintenance
Offer reasonable asynchronous performance
A manageable torque-speed curve
Stable operation under load
Generally satisfactory efficiency
Range in size from few Watts to several MW
Introduction (cont’d)
Some disadvantages of induction motors
are:



Speeds not as easily controlled as dc motors
Draw large starting currents, typically 6-8 x
their full load values
Operate with a poor lagging power factor
when lightly loaded
Introduction (cont’d)
New designs for high performance
induction machines, such as in high
speed motors for gas compressors, will
be required to have new characteristics
from existing machines, it is important to
have a good fundamental understanding
of these types of machines.
Goal: To develop a “simple” model for the
induction machine that is useful for
control and simulation.
Structure of an Induction Machine
Two types of induction machine:
Wound rotor or squirrel cage rotor
Rotating Magnetic Field and Slip
We previously showed that a balanced set
of three-phase currents flowing in a set of
symmetrically placed, three-phase stator
windings produces a rotating mmf given by:
34N
F ( , t ) 
I m cos( ae   et )
2 P
e
a
[eq. (6.1) Ong, eq. (2.9) Bose]
where ae is the electrical angle measured
from the a-phase axis and e is the angular
speed of the stator mmf in electrical
radians/second.
Rotating Magnetic Field and Slip
(cont’d)
Rotating Magnetic Field and Slip
(cont’d)
In mechanical radians/sec. the synchronous
speed is related to the electrical speed by:
 sm
2
 e
P
If the rotor is rotating at an angular speed rm
the slip speed is simply equal to sm - rm.
The “slip”,s, is the normalized slip speed and
is given by:
 sm   rm e   r
s

 sm
e
Torque Production
The torque produced by an induction motor
may be derived and expressed by the
following equation: (see ref. [1] in Bose)
P
Te     lrBp Fp sin 
2
where P= # of poles
l = axial length of motor
r = radius of motor
Bp= peak air-gap flux density
Fp= peak value of rotor mmf

and     r
2
Per-Phase Equivalent Circuit Model
A per-phase transformer-like equivalent
circuit is shown below:
Per-Phase Equivalent Circuit
Model (cont’d)
Synchronously rotating air gap flux wave
generates a counter emf Vm. This in turn is
converted to a slip voltage in the rotor
phase, Vr’ = nsVm, where n=rotor:stator
turns ratio, and s=normalized slip.
Stator terminal voltage, Vs = Vm + VRs +VLls
where VRs=voltage drop across stator
resistance (Rs) and VLls=voltage drop across
stator leakage inductance (Lls).
Per-Phase Equivalent Circuit
Model (cont’d)
Excitation current, I0 = Ic + Im
where Ic is core loss current (=Vm/Rm)
and Im is magnetizing current (=Vm/e Lm )
Rotor-induced voltage, Vr’ = VRr’ + VLl’
where VRr’ = voltage drop across rotor resistance
and VLl’ = voltage drop across rotor leakage
inductance
The induced voltage in the rotor leads to a rotor
current Ir’ at slip frequency sl.
Per-Phase Equivalent Circuit
Model (cont’d)
The stator current, IS = I0 + Ir
where Ir is the rotor-reflected current
induced in the stator.
I0
Per-Phase Equivalent Circuit
Model (cont’d)
2
n
sVm
Vm
'
I r  nI r  '

'
Rr  j sl Llr  Rr 
   j e Llr
 s 
Rr'
Rr  2
n
L'lr
Llr  2
n
Per-Phase Equivalent Circuit
Model (cont’d)
Per-Phase Equivalent Circuit
Model (cont’d)
Torque expression can be written as:
3 P ˆ
Te   ˆ m I r sin 
2 2 
where ˆ m = peak value of air gap flux
linkage/pole
and Iˆr = peak value of rotor current
Per-Phase Equivalent Circuit
Model: Power Expressions
Input Power: Pin  3Vs I s cos 
where cos is input PF
Stator copper loss: Pls  3I s2 Rs
Rotor copper loss:
Core loss:
Plr  3I Rr
2
r
Plc  3Vm2 / Rm
Power across air gap: Pg  3I r2 Rr / s
Output power: Po  Pg  Plr  3Ir2 Rr (1 s / s)
Shaft Power: Psh  Po  PFw
where PFw is friction and
windage power loss
Per-Phase Equivalent Circuit
Model: Torque Expression
The torque can be expressed as:
1 s
 P  2 Rr
Te 

I Rr
 3  Ir
m m
s
 2  se
Po
3
2
r
2
2
where m    r    (1  s)e is the rotor
P
P
mechanical speed (radians/sec.)
Per-Phase Equivalent Circuit
Model: Torque Expression (cont’d)
Using a little algebra (see Bose) it can
be shown that the torque may be further
expressed as:
P
Te  3   Lm I m I a
2
where I a  I r sin  .
This torque expression is similar to that for
a dc motor, where Im = magnetizing
component of stator current and Ia =
armature component of stator current.
Simplified Per-Phase Equivalent
Circuit
A simplified circuit dropping Rm and
shifting Lm to the input is applicable to
integral horsepower machines.
Performance of this equivalent circuit is
typically within 5%.
Simplified Per-Phase Equivalent
Circuit Model (cont’d)
The current Ir in this circuit is given by:
Ir 
Vs
( Rs  Rr / s)2  e2 ( Lls  Llr )2
The torque of the motor using this circuit
is given by:
Vs2
 P  Rr
Te  3  
2
2
2
2
s

(
R

R
/
s
)


(
L

L
)
  e s
r
e
ls
lr
Example of Calculating Efficiency
of an Induction Motor
Example 5.1 Krishnan
Flowchart for Computing Steady
State Performance of Induction Motor
Ref: R. Krishnan,
“Electric Motor Drives”
Torque-Speed Curve of Induction
Motor
The torque-speed curve as a function of
slip can be calculated from the equation
two slides back.
Torque-Speed Curve of Induction
Motor (cont’d)
Three regions in torque-speed curve:
1) Plugging (braking) region (1<s<2)
Rotor rotates opposite to direction of
air gap flux. Can happen, for example,
if stator supply phase sequence
reversed while rotor is moving.
2) Motoring region (0<s<1)
Te=0 at s=0. As s increases (speed
decreases), Te increases until max.
torque (breakdown torque) is reached.
Beyond this point, Te decreases with
increasing s.
Torque-Speed Curve of Induction
Motor (cont’d)
3) Regenerating Region (s<0)
Here the induction machine acts as a
generator. Rotor moves faster than air
gap flux resulting in negative slip.
Torque-Speed Curve of Induction
Motor (cont’d)
Ref: R. Krishnan,
“Electric Motor Drives”
Performance Characteristics of
Induction Motor
Ref: R. Krishnan, “Electric Motor Drives”
Starting Torque of Induction Motor
The starting torque of an induction motor
is given by substituting for s=1 and is
given by:
Vs2
 P  Rr
Te  3  
2
2
2
2

(
R

R
)


(
L

L
)
  e s
r
e
ls
lr
Starting Torque of Induction Motor
(cont’d)
This torque can be enhanced for line start
motors (ones started directly with full line
voltage) by increasing the rotor resistance.
This can be achieved by connecting
external resistors in the case of slip ring
rotors. However, with squirrel cage rotors
where the rotor is shorted, deep bar or
double-cage rotors can be used to increase
starting torque.
Characterizing Induction Motors
One way to characterize an induction
motor is with the No-load/blocked rotor
tests which yield the per-phase equivalent
circuit model shown earlier (see figure
below).
ias
vas
iar
=M
Characterizing Induction Motors
(cont’d)
We can characterize an induction motor
with the variables Rs, Lls, M, Llr, Rr
determined through lab tests using
balanced 3 excitation. This circuit
described the impedance perceived per
phase on a line-neutral connected machine.
Everything in the dashed box is a rotor
quantity that has been “referred” to the
stator by the ideal transformer in the
machine model. From now on, assume that
Llr, Rr and iar are referred to the stator.
Characterizing Induction Motors
(cont’d)

No-Load Test (s=0)
Equivalent circuit:
Ref: R. Krishnan, “Electric Motor Drives”
Characterizing Induction Motors
(cont’d)

No-Load Test (s=0) yields:
In sinusoidal steady state, ignoring
resistances:
vas  ias X s  ibs X ab  ics X ab
Las
But -ias = (ibs+ics)
3 

 vas  ias [ X s  X ab ]  ias Lls  Lsr

2 
From transformer model:
vas  ias  Lls  M 
=>
3
M  Lsr
2
Characterizing Induction Motors
(cont’d)

Blocked rotor test (s=1) yields
estimates of Lls and Llr. Equivalent
circuit at standstill is shown below:
Ref: R. Krishnan, “Electric Motor Drives”
Characterizing Induction Motors
(cont’d)

Ohmmeter/Power loss tests give Rs
and Rr.
So, with Llr, Rr and all ir’s understood as
referred rotor quantities, the “stator-side”
tests identify all the model parameters for
the induction motor.
Example of Determining Induction
Motor Model Parameters
Example 5.2 Krishnan
NEMA Classification of Induction
Motors
The National Electrical Manufacturers
Association (NEMA) has classified
induction motors based on their
torque-slip characteristics. (see text
for details)
Circuit Model of a Three-Phase Induction
Machine (State-Space Approach)
Voltage Equations
Stator Voltage Equations:
d as
vas  ias rs 
dt
d bs
vbs  ibs rs 
dt
d cs
vcs  ics rs 
dt
Voltage Equations (cont’d)
Rotor Voltage Equations:
d ar
var  iar rr 
dt
d br
vbr  ibr rr 
dt
d cr
vcr  icr rr 
dt
Flux Linkage Equations
Model of Induction Motor
To build up our simulation equation, we
could just differentiate each expression
for , e.g.
d as d [first row of matrix]
vas 

dt
dt
But since Lsr depends on position,
which will generally be a function of
time, the trig. terms will lead to a mess!
Park’s transform to the rescue!