Transcript Teknologi Elektrik

Performance Characteristics
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The equivalent circuits can be used to predict the
performance characteristics of the induction machine.
The important performance characteristics in the steady
state are:
- efficiency
- power factor
- stator current
- starting torque
- maximum torque (pull-out), etc
I th  I '2 
Vth
2


R
'

  Rth  2    X th  X '2 2 


s 


1
2
Performance Characteristics
Eqn. X
Performance Characteristics
*eqn. X
At high
slip
At low
slip
Rth 
• At low slip, torque
proportional to slip s
•At high slip, torque
inverse proportional to slip
&
R '2
 X th  X 1
s
R '2
 Rth
s

Note if approximate circuit is used to get the
equation of torque, then
Since L m is large, thecircuit abovecan be drawn
I2 
V1
2
R2 

2
2
 R1    syn L1  L2 
s 

P ower at therotor(per phase),
2 R 
Po  Pag  I 2  2 
 s 
Electromagnet ic(developed) torque,
Te 
3Po
syn
3V1
2
R2

2

 s
R2 
2
2
syn  R1    syn L1  L2  
s 


T – N Single frequency characteristic
T ORQUE(+)
e
m
P LUGGING
MOT ORING
Tem
(max t orque or
pull-out t orque)
e
e
m
m
GENERATING
Tes (starting t orque)
-Ns
0 unit y slip
(standstill)
2
zero slip
 e (sync.speed)
rat ed slip
SP EED
SLIP ,s
T ORQUE(-)
Note that:
syn  m
s
; at standsill s  1, at syncspeed, s  0.
syn
• Select one frequency ()
•Select V1
• Varies s ( at particular s,
get T)
•Repeat for other s to get T
T – N Single frequency characteristic
CURRENT
TORQUE
operating point
(rated torque)
EFFICIENCY
POWER
FACTOR
rated current
rated slip
SLIP
1.0
Standstill
0
synchronous speed
T – N Single frequency characteristic
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As slip is increased from zero (synchronous), the
torque rapidly reaches the maximum. Then it
decreases to standstill when the slip is unity.
At synchronous speed, torque is almost zero.
At standstill, torque is not too high, but the current is
very high. Thus the VA requirement of the IM is
several times than the full load. Not economic to
operate at this condition.
Only at “low slip”, the motor current is low and
efficiency and power factor are high.
Maximum Torque
Differentiate eqn.
dT/ds, and equate to
zero
If R1
small
Increase
R2,
increase
slip max,
increase
staring
torque
If R1
small
Varying R2
Maximum Torque
• Maximum air gap
power transfer occurs
at impedance
matching principle –
Another approaches

R2 '
2
 Rth  ( X th  X 2 ' ) 2
Sm
R ext

1
2
• Rext to be added to
produce Tmax at
starting, ie at s = 1 is

R2 ' Rext
2
 Rth  ( X th  X 2 ' ) 2
Sm

1
2
Current and Power Factor
Stator current vs. speed
Power factor vs. speed
Efficiency
(1-s)Pag
sPag
Efficiency vs. speed
Internal efficiency
*To get Max. efficiency, s
must be very low
Power Flow
0<s<1
0<s
s>0
Example 4*
A three-phase 460 V, 1740 rpm, 60 Hz 4-pole wound rotor induction star
connected motor has the following parameter/phase:
R1 = 0.25  , R2’ = 0.2  , X1 = X2’= 0.5  , Xm = 30 
The rotational losses are 1700 W. With the rotor terminal short circuited,
find:
a) i) Starting current when started on full load
ii) Starting torque
b) i) Full load slip ii) Full-load current iii) Full-load power factor
iv) Ratio of starting current to full load current v) Full-load torque
vi) Internal efficiency and motor efficiency at full load
c) i) Slip at maximum torque ii) Maximum Torque
d) How much external resistance/phase should be connected in the rotor
circuit so that maximum torque occurs at start?
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Sol- pg13
Sen 241
Example 5
A three-phase 460 V, 60 Hz 6 -pole wound rotor
induction motor drives a constant load of 100 N-m at
speed of 1140 rpm when the rotor terminal is shortcircuited. It requires to reduce speed to 1000 rpm by
inserting resistance in rotor circuit.
Determine the value of resistance if the rotor winding
resistance / phase is 0.2 ohms. Neglect rotational
losses. The stator to rotor turn ratio is unity.
Since the
l
R2 +R2ext
R2
Sol_pg14
TL
N2
Sen 244
N1
developed
torque Tm =
load torque
TL
R2 R2  Rext

S1
S2
Wound Rotor
By changing the
impedance (R)
connected to the rotor
circuit, the speed/current
and speed/torque curves
can be altered.
Used primarily to start a high
inertia load or a load that
requires a very high starting
torque across the full speed
range with relatively low
current from zero speed to full
speed
Slip ring
15
Example 6
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The following test results are obtained from a three-phase, 100hp,
460 V, eight-pole, star connected squirrel-cage induction machine.
No load test:
460 V, 60 Hz, 40 A, 4.2 kW
Blocked-rotor test: 100 V, 60 Hz, 140 A, 8 kW
Average dc resistance between two stator terminals is 0.152 Ω.
(a) Determine the parameters of the equivalent circuit.
(0.076 Ω, 0.195 Ω, 6.386 Ω, 0.195 Ω, 0.062 Ω).
(b) The motor is connected to a three-phase , 460 V, 60 Hz supply and
runs at 873 rpm. Determine the input current, input power, air gap
power, rotor copper loss, mechanical power developed, output
power, and efficiency of the motor.
( 127.9/-27o A, 90.82 kW, 87.09 kW, 2.613 kW, 84.48 kW, 80.64 kW,
88.79 %)
Sol_pg16
Sen 282 (pb 5.6)
Classes of Squirrel-Cage Motor
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To meet the various starting and running requirements of
a variety of industrial applications, several standard ( T
vs. N) designs of squirrel-cage motors are available from
manufacturer’s stock.
The most significant design variable in these motors is
the effective resistance of the rotor cage circuit ( for
wound rotor)
Class A Motors
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Characterized by normal starting torque, high starting
current and low operating slip.
Low rotor circuit resistance and therefore operate
efficiently with a low slip (0.005<s<0.015) at full load.
Suitable for applications where the load torque is low at
start (such as fan or pump) so that full speed is achieved
rapidly, thereby eliminating the problem of overheating
during starting.
In larger machines, low voltage starting is required to
limit the starting current.
Class B Motors
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Characterized by normal starting torque, low starting
current and low operating slip.
The starting current is about 75 % of that for class A.
The starting current is reduced by designing for relatively
high leakage reactance by using either deep-bar rotors
or double- cage rotors.
The full load slip and efficiency are as good as those of
class A motors.
Good general-purpose motors and have a wide variety of
industrial applications. Suitable for constant speed drives.
Examples are drives for fans, pumps, blowers, and
motor-generator sets.
Class C Motors
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Characterized by high starting torque and low starting
current.
A double-cage rotor is used with higher rotor resistance
than is found in class B motors.
The full-load slip is somewhat higher and the efficiency
lower than for class A and class B motors.
Class C motors are suitable for driving compressors,
conveyors, crushers, and so forth.
Class D Motors
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Characterized by high starting torque, low starting current and
high operating slip.
The torque-speed characteristic is similar to that of a wound-rotor
motor with some external resistance connected to the rotor circuit.
The full-load operating slip is high (8 to 15 %), and therefore the
running efficiency is slow.
The high losses in the rotor circuit require that the machine be large
(and hence expensive) for a given power.
Suitable for driving intermittent loads requiring rapid acceleration
and high impact loads.
In the case of impact loads, a flywheel is fitted to the system which
delivers some of its kinetic energy during the impact.
Speed Control
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Pole Changing
Line Voltage Control
Line Frequency Control
Constant-slip Frequency Operation
Closed-loop Control
Constant-Flux Operation
Constant-current Operation
Rotor Resistance Control
Rotor Slip Energy Recovery
Speed control of induction machine
Given a load T–N characteristic,
the steady-state speed can be changed
by altering the profile of T–N of the motor:
Rotor Resistance control
Pole changing
Synchronous speed change with changes
number of poles (change the stator
winding/coil connection)
Discrete step change in speed/ expensive
Variable line voltage (amplitude), frequency
fixed
Torque  V2
E.g. using 3-phase autotransformer (variac) or
solid state controller
Slip becomes high as voltage reduced – low
efficiency
For wound rotor only
Variable line voltage (amplitude),
variable frequency
. Most popular method
. Using power electronics converter
. Operated at low slip frequency
Variable line voltage, fixed frequency
600
V= 1pu
500
Torque
400
V= 0.71pu
300
200
Fan (TL)
load
V= 0.25pu
100
0
0
20
40
60
80
w (rad/s)
100
120
140
160
Auto Transformer Voltage Control
Solid State Voltage Control
Closed Loop Operation Voltage Control
Rotor resistance Control
Open Loop Control Scheme
Closed Loop Control Scheme
Typical IM Drive System - Variable voltage, variable
frequency
+

IDC
+
VDC
IM

Modulation Index,
Control both V and freq, f
n
Supply Rectifier and Filter
3-phase pwm Inverter
PWM Inverter
IM
Variable voltage, variable frequency
Constant V/f
900
800
50Hz
700
30Hz
Torque
600
E V
  m
f f
500
10Hz
400
300
200
100
0
0
20
40
60
80
100
120
140
160
VVVF, Constant V/f – open-loop
VSI
Rectifier
3phase
supply
IM
C
f
Ramp
s
*
+
V
Pulse
Width
Modulator
Example Final
Question 4
(a) Explain briefly three methods for controlling the speed of an
induction motor. (6 marks)
(b) Draw a typical torque-speed characteristic of an induction
motor and label key quantities. (3 marks)
(c) A three-phase, 415 V, 1450 rpm, 50 Hz, four-pole wound-rotor
induction motor has the following parameters per phase:
R1 = 0.25 , R2’ = 0.2 
X1 = X2’ = 0.5 , Xm = 30 
The rotational losses are 1700 W. With the rotor terminals shortcircuited, determine:
(i) Starting current when started direct on full voltage. (4 marks)
(ii) Starting torque. (4 marks)
(iii) Full-load current. (4 marks)
SEMESTER 1
(iv) Full-load torque. (4 marks)
SESI 2007/2008
eg 5.4 pg 241
Example
QUESTION 4
(a) Explain the working principle of a three-phase induction machine on the basis of
magnetic fields.
(b)Show through a power flow diagram, how electrical power input is converted into
mechanical power output in an induction motor.
(c) (c) A 3 phase , 415 V, 1450 rpm, 50 Hz, four-pole wound rotor induction motor has the
following Thevenin’s equivalent circuit parameters per phase:
Vth = 236 V
Rth = 0.25 W
Xth = 0.5 W
X2 = 0.5 W R2’ = 0.2 W
The motor drives a constant load of 100 Nm at rated speed when the rotor terminals are
short-circuited. Neglect rotational losses.
(i) Draw the Thevenin’s equivalent circuit of induction machine.
(ii) How much external resistance per phase should be connected in the rotor circuit so that
maximum torque occurs at start-up?
(iii) It is required to reduce the speed of the motor to 1400 rpm by inserting resistance in the
rotor circuit. How much external resistance per phase should be connected in the rotor
circuit?
(iv)Draw torque-speed characteristics of the motor and load to show the conditions in (iii)
with and without external rotor resistance.
SEMESTER 1
SESI 2008/2009
Sol_pg31