induction motor speed control
TRANSCRIPT
Different Speed control methods for Induction motor
• By changing rotor resistance (only possible in slip-ring induction motor (SRIM))
•By changing the applied voltage and frequency to the induction motor, while keeping the
ratio constant.
• By changing stator poles by reconnecting the stator coils
•By changing only the applied voltage to stator
•By recovering the slip power (only possible in SRIM)
Equivalent Circuit Modification
V1
I1
R1
jX1 jX2’
R2’
((1-s)/s)R2’
jXm
I2’
Im
Assuming stator drop is negligible (this is true except for
low speed operation when the stator resistance drop
becomes comparable to supply voltage) the equivalent
circuit of the induction motor can be modified to the
bottom figure from the top figure.
V1
R1
jX1 jX2’
R2’
((1-s)/s)R2’
jXm
I2’
Im
Output Torque
Ouput torque output neglecting rotational losses
= 𝑇𝑜𝑢𝑡 = 𝑃𝑜𝑢𝑡
𝜔=
𝑃𝑜𝑢𝑡
(1 − 𝑠)𝜔1=
𝑃𝑎𝑔
𝜔1= 𝐼2
′ 2𝑅2
′
𝑠𝜔1
From the equivalent circuit,
𝐼2′ =
𝑉1
𝑅1 +𝑅2
′
𝑠 + 𝑗 𝑋1 + 𝑋2
′
𝐼2′ =
|𝑉1|
𝑅1 +𝑅2
′
𝑠
2
+ 𝑋1 + 𝑋2′
2
Substituting the magnitude of current in torque expression,
𝑇𝑜𝑢𝑡 =|𝑉1|2
𝑅1 +𝑅2
′
𝑠
2
+ 𝑋1 + 𝑋2′
2
∙𝑅2
′
𝑠𝜔1
V1
R1
jX1 jX2’
R2’
((1-s)/s)R2’
jXm
I2’
Im
Maximum Torque 𝑇𝑜𝑢𝑡 =
|𝑉1|2
𝑅1 +𝑅2
′
𝑠
2
+ 𝑋1 + 𝑋2′
2
∙𝑅2
′
𝑠𝜔1
To obtain the maximum torque, the above expression of torque is differentiated with respect to
slip and equated to zero.
𝑇𝑚𝑎𝑥 =|𝑉1|2
𝑅1+ 𝑅12+ 𝑋1+𝑋2
′ 2
∙1
2𝜔1 ≈
|𝑉1|2
2𝜔1 𝑋1+𝑋2′
(neglecting 𝑅1)
Slip at which maximum torque occurs
𝑠𝑇𝑚𝑎𝑥 =𝑅2′
𝑋1+𝑋2′
This shows that while the maximum torque is independent of rotor resistance, the slip at which
the maximum torque occurs is dependent on rotor resistance.
Speed control by changing rotor resistance in SRIM
In case of a slip ring (wound rotor) induction motor
(SRIM) the speed can therefore be varied by changing the
rotor resistance. More on this will be discussed later.
s = 1 slip
T
Tmax
s = 0
TL
Increasing R2
Changing Tm profile with fixed
peak torque by changing R2
Decreasing speed with
increasing R2
Speed control by keeping constant flux (ratio of
applied voltage and frequency)
slip s = 0
s = 1
T
Tmax
Decreasing f
Changing Tm profile with fixed peak toruqe
with changing frequency and voltage but
with their ratio constant
Decreasing speed for decreasing voltage
and frequency but with their ratio constant
A different strategy is employed to achieve speed control for squirrel cage induction motor. It is
applicable for slip ring induction motor as well.
𝑇𝑚𝑎𝑥 ≈|𝑉1|2
2𝜔1 𝑋1 + 𝑋2′
𝜔1 = 2𝜋𝑓/(𝑝) where p is the pole pair number.
Also, 𝑋1 = 2𝜋𝑓 𝐿1 and 𝑋2′ = 2𝜋𝑓 𝐿2
′
Therefore, 𝑇𝑚𝑎𝑥 =𝑝|𝑉1|2
8𝜋2 𝐿1+𝐿2′ 𝑓2
=𝑝
𝑉1 𝑓
2
8𝜋2 𝐿1+𝐿2′
Thus if the ratio of supply voltage to frequency , i.e., 𝑉1 𝑓 maintained constant,
the peak flux of the motor (essentially the magnetizing current) and hence the maximum torque
can be held constant.
Speed control of Induction motor: Pole changing
By changing poles the synchronous speed and hence the motor operating speed can be changed.
Poles can be changed by changing coil connections of the stator. Normally, poles can be changed
in the ratio of 2:1. The speed can be changed only in discrete steps. Hence it is not very popular.
The figure above shows how to reconnect a 4 pole motor as a 2 pole one.
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++
-
+
-
+
-
--
+
+
N
S
+
+
+
+
+
+
-
-
--
-
-
NS
NS
Speed control of Induction Motor: By varying stator supply voltage only
By using ac voltage controller as shown above (using back-to-back connected SCRs )
the three phase stator supply voltage can be smoothly controlled. Speed also
can be reversed by using the connections shown by dotted lines, with the other
back-to-back connected SCRs left ungated.
However, since the torque depends on the square of the voltage, the output torque
is reduced greatly when the voltage alone is reduced as can be easily understood from
the following expression of maximum torque
Speed control of Induction Motor: By varying rotor resistance in a SRIM
Example:
A 3-phase, 460V, 60 Hz, 6 pole SRIM drives a constant load torque of 100 N-m at a speed of
1140 rpm when the rotor terminals are short circuited. It is required to reduce the speed of the
motor to 1000 rpm by inserting resistances in the rotor circuit. Determine the value of the
external resistance if 𝑅2, the effective rotor winding resistance is 0.2Ω. The stator to rotor turns
ratio is 1:1.
Solution :
𝑁1 = 120𝑓
𝑃=
120∗60
6= 1200 rpm.
∴ slip at 1140 rpm = 𝑠1 =1200 − 1140
1200= 0.05
∴ slip at 1000 rpm = 𝑠2 =1200 − 1000
1200= 0.167
Y
R
B
Motor
Example on SRIM speed control by varying rotor resistance
Now
𝑇𝑜𝑢𝑡 =|𝑉1|2
𝑅1 +𝑅2
′
𝑠
2
+ 𝑋1 + 𝑋2′
2
∙𝑅2
′
𝑠𝜔1
We can keep the speed constant by keeping 𝑅2′
𝑠 constant.
∴𝑅2
′
𝑠1=
𝑅2′ + 𝑅𝑒𝑥𝑡
′
𝑠2
Since stator to rotor turns ratio is 1:1,
∴𝑅2
𝑠1=
𝑅2 + 𝑅𝑒𝑥𝑡
𝑠2
or 0.2
0.05=
0.2 + 𝑅𝑒𝑥𝑡
0.167
∴ 𝑅𝑒𝑥𝑡 =0.2 ∗ 0.167
0.05− 0.2 = 0.468Ω.
Electronic control of Rotor resistance in a SRIM
B
R
Y
Motor
RM
By varying the duty cycle ‘d’ of the MOSFET M the effective rotor resistance can
be changed by 0.5(1-d)R per phase.
Slip Power Recovery Scheme
C1
C2
Slip Power Recovery Scheme(2)
Instead of dissipating power in resistance (external resistance added in the rotor circuit), power
can be returned using a phase controlled rectifier.
Power delivered back to the source is 𝑉𝑑𝑐 𝐼𝑑𝑐
Effective resistance added to the rotor is given by 𝑅𝑒𝑓𝑓 = 𝑉𝑑𝑐
𝐼𝑑𝑐
Assuming that the rotor resistance𝑅2 is negligible compared 𝑅𝑒𝑓𝑓 , the entire slip power is fed
back to the source.
𝑠𝑃𝑎𝑔 = 𝑉𝑑𝑐 𝐼𝑑𝑐
For a three phase diode bridge rectifier, the average voltage is given by the expression
𝑉𝑑𝑐 =3 2
𝜋𝑠𝑎1𝑉1 = 1.35𝑠 𝑎1𝑉1, where 𝑎1 is the motor rotor to stator turns ratio and 𝑉1 is the
stator line-line voltage.
Slip Power Recovery Scheme(3)
For a three phase fully controlled SCR bridge converter, the average voltage is given by
𝐸𝑎𝑣 =3 2
𝜋𝑎2𝑉1 cos 𝛼 = 1.35𝑎2𝑉1 cos 𝛼 , where 𝑎2 is the transformer secondary to primary
turns ratio.
Since under steady state 𝑉𝑑𝑐 = 𝐸𝑎𝑣
∴ 𝑠𝑎1 = 𝑎2 cos 𝛼
∴ 𝑠 =𝑎2
𝑎1| cos 𝛼 |
Thus the speed of the motor can be controlled by changing the firing delay angle of the converter.
Also
𝑠𝑃𝑎𝑔 = 𝑉𝑑𝑐 𝐼𝑑𝑐
or 𝑠𝑇𝜔1 = 𝑉𝑑𝑐 𝐼𝑑𝑐 = 1.35𝑠𝑎1𝑉1 𝐼𝑑𝑐
or 𝑇 =1.35
𝜔1𝑎1𝑉1 𝐼𝑑𝑐
or 𝑇 ∝ 𝐼𝑑𝑐 .
Slip Power Recovery Scheme( Example) An 8 pole, 440V, three phase, 60 Hz, slip ring induction motor is being used for slip power
recovery scheme. The stator to rotor turns ratio is 1:0.75. The stator of the motor is connected
in delta, while the rotor is connected in star. The power is fed back to the utility supply
through a star-delta transformer after the inverter, with the star side (primary) connected to the
460V utility supply. The primary to secondary turns ratio for the transformer is 2:1. Calculate
a) the power sent back to the utility supply, b and b) the speed of the motor and c) the input
power drawn by the motor if the dc link current is 200A.𝛼 = 135°.
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+
-
+
+
-
Motor
Three Phase Transformer
440
330√
3s
440*0.75*s
440/√3
220/
√3
Vdc
Eav
R
Y
B
Idc
Slip Power Recovery Scheme( Example)(2)
𝑉𝑑𝑐 = 1.35 ∗ 0.75 ∗ 440 ∗ 3 ∗ 𝑠 = 1.35 ∗ 330 3 ∗ 𝑠 = 771.63𝑠 V
𝐸𝑎𝑣 = 1.35 ∗220
3∗ cos 135° = 1.35 ∗ 89.8 = 121.25 V
∵ 𝑉𝑑𝑐 = 𝐸𝑎𝑣
∴ 𝑠 =121.25
771.63= 0.1571
∴ Power sent back to the source = 𝑉𝑑𝑐 𝐼𝑑𝑐 = 121.25 ∗ 200 W = 24.25 kW.
𝑇 =𝑉𝑑𝑐 𝐼𝑑𝑐
𝑠𝜔1=
24250
0.1571∗2𝜋∗900
60
= 1637.45 Nm.(∵ 𝑁1 =120𝑓
𝑃=
120∗60
8= 900 RPM)
Now Motor output power = 𝑉𝑑𝑐 𝐼𝑑𝑐 ∗(1−𝑠)
𝑠= 24250 ∗
1−0.1571
0.1571= 130.11 kW.
∴ Input power to the motor = Motor output power + Power sent back to the source
= 130.11 + 24.25 kW = 154.36 kW.
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+
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+
+
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Motor
Three Phase Transformer
440
33
0√
3s
440*0.75*s
440/√3
220/
√3
Vdc
Eav
R
Y
B
Idc
Speed Control Scheme Using Slip Power Recovery
Transformer
Tachometer
Speed
Controller+
-+
-ωref
ω
Vd Vi
Id
SRIM
R
Y
B
aFiring
Angle
Controller
LimiterLimiter
Mechanical
load
Id,ref
Scalar v/f method of speed control of induction motor using a voltage source inverter(VSI) :
(i) Below Base Speed
V1
R1
jX1 jX2’
R2’
((1-s)/s)R2’
jXm
I2’
Im
𝑇𝑜𝑢𝑡 =|𝑉1|2
𝑅1+𝑅2′
𝑠
2
+ 𝑋1+𝑋2′
2
∙𝑅2′
𝑠𝜔1
Assuming slip to be small,
𝑅2′
𝑠≫ 𝑅1 &
𝑅2′
𝑠 ≫ 𝑋1 + 𝑋2
′
∴ 𝑇𝑜𝑢𝑡 ≈|𝑉1|2
𝑅2′
𝑠
2
∙𝑅2′
𝑠𝜔1=
|𝑉1|2
𝑅2′ ∙
𝑠
𝜔1
Multiplying and dividing by 𝜔1
𝑇𝑜𝑢𝑡 ≈ |𝑉1|
𝜔1
2
∙𝑠𝜔1
𝑅2′
Scalar v/f method of speed control of induction motor using a voltage source inverter(VSI) (2)
From the above equation it is clear that if the ratio of voltage to supply frequency is maintained
constant, then torque is proportional to slip speed 𝑠𝜔1. The approximation is valid as long as the
stator resistance 𝑅1 ≪ 𝑋1 + 𝑋2′ so that the maximum torque capability of the motor given by
𝑇𝑚𝑎𝑥 =|𝑉1|2
𝑅1+ 𝑅12+ 𝑋1+𝑋2
′ 2
∙1
2𝜔1 ≈
|𝑉1|2
2𝜔1 𝑋1+𝑋2′
=𝑝 𝑉1 𝑓 2
8𝜋2 𝐿1+𝐿2′ is not severely compromised.
For low frequency operation, 𝑉1 has to be boosted in order to accommodate for the increasing
influence of 𝑅1. The effect of not boosting 𝑉1 can lead to drastic reduction of peak torque as
shown by the figure below.
𝑇𝑜𝑢𝑡 ≈
|𝑉1|
𝜔1
2
∙𝑠𝜔1
𝑅2′
Scalar v/f control using VSI: Example (1) A three phase, induction motor drives a constant torque over a 4:1 speed range by frequency
control with the ratio of 𝑉1 𝑓 maintained constant. If the motor’s reactance to stator resistance
ratio is 5 at rated frequency of operation, calculate the effect on the peak torque of operation at
the lowest speed. What value of 𝑉1 𝑓 is required at the lowest speed if the peak torque is to
remain constant? Solution:
At rated frequency,
𝑇𝑚𝑎𝑥 =|𝑉1|2
𝑅1+ 𝑅12+ 𝑋1+𝑋2
′ 2
∙1
2𝜔1=
|𝑉1|2
𝑅1+𝑅1 1+ 𝑋1+𝑋2′ 𝑅1
2
∙1
2𝜔1=
|𝑉1|2
𝑅1+𝑅1 1+ 5 2 ∙
1
2𝜔1 Nm
or 𝑇𝑚𝑎𝑥 =1
1+ 1+ 5 2
|𝑉1|2
2𝑅1𝜔1= 0.082
|𝑉1|2
𝑅1𝜔1 Nm
At 1/4th
rated frequency, if the voltage is also reduced to one-fourth of its rated value then
maximum torque developed is
𝑇𝑚𝑎𝑥 = 𝑉1 4 2
𝑅1+ 𝑅12+ 𝑋1+𝑋2
′ 2
∙1
2 𝜔1
4
= 𝑉1 4 2
𝑅1+𝑅1 1+ 𝑋1+𝑋2′ 4𝑅1
2
∙2
𝜔1=
𝑉1 4 2
𝑅1+𝑅1 1+ 5/4 2 ∙
2
𝜔1 Nm
or 𝑇𝑚𝑎𝑥 =1/8
1+ 1+ 1.25 2
|𝑉1|2
𝑅1𝜔1= 0.0481
|𝑉1|2
𝑅1𝜔1 Nm
Scalar v/f control using VSI: Example (1) Contd...
Thus the maximum torque developed has gone down at lowest speed by about 58.5%. In order to
compensate for the maximum torque the supply voltage has to be boosted. This can be estimated
from the following:
𝑇𝑚𝑎𝑥 = 𝑘 𝑉1 2
𝑅1+ 𝑅12+ 𝑋1+𝑋2
′ 2
∙1
2 𝜔1
4
= 𝑘 𝑉1 2
𝑅1+𝑅1 1+ 𝑋1+𝑋2′ 4𝑅1
2
∙2
𝜔1=
𝑘 𝑉1 2
𝑅1+𝑅1 1+ 5/4 2 ∙
2
𝜔1 Nm
or 𝑇𝑚𝑎𝑥 =2𝑘2
1+ 1+ 1.25 2
|𝑉1|2
𝑅1𝜔1= 0.769𝑘2 |𝑉1|2
𝑅1𝜔1= 0.082
|𝑉1|2
𝑅1𝜔1 .
𝑘 = 0.082
0.769 =0.326
Thus instead of 25%, 32.6% of the of rated voltage is to be applied if the maximum developed
torque has to remain constant at 1/4th
rated speed.
Scalar v/f control using VSI: Example (2) A 450 V, 50 Hz, 1450 rpm, 25 kW, Y connected, 3 phase induction motor delivers rated torque
at all speed. The motor equivalent circuit parameters at rated frequency are 𝑅1 = 0.1 Ω, 𝑅2′ =
0.17 Ω, 𝑋1 = 0.3 Ω, 𝑋2′ = 0.5 Ω, 𝑋𝑚 = 23.6 Ω. The machine flux is kept constant at rated value
under all operating conditions. Calculate the line current, power-factor, efficiency at 1/5th
the
rated speed.
1/5th
the rated speed =1450 5 = 290 rpm.
Since torque is constant at rated value under all speed, slip speed or slip frequency will have to
be the same at the rated value.
Now synchronous speed 𝑁1 =120𝑓
𝑃=
120∗50
𝑃> 1450 rpm.
∴ 𝑃 <120 ∗ 50
1450< 4.13
Since 4 is the closest and lesser even number to 4.13, 𝑃 = 4.
∴ 𝑁1 =120 ∗ 50
4= 1500 rpm.
Thus slip speed =𝑁1 − 𝑁 = 1500 − 1450 = 50 rpm.
Scalar v/f control using VSI: Example (2) Contd...
Thus synchronous speed at 1/5th
rated speed 𝑁1/5 = 290 + 50 = 340 rpm.
Therefore slip at 1/5th
rated speed =50
340= 0.1471.
Since the motor delivers rated torque at 1/5th
rated speed, therefore it delivers 1/5th
rated power at
this speed.
∴ 3 𝐼2′ 2 𝑅2
′ (1−𝑠)
𝑠= 3 𝐼2
′ 2 0.17(1−0.1471 )
(0.1471 )=
25000
5= 5000 W.
∴ 𝐼2′ =
5000 ∗ 0.1471
0.51 ∗ 0.8529= 41.12 A
Also at 1/5th
rated speed the new values of 𝑋1 = 0.3 340
1500 = 0.068 Ω , 𝑋2
′ = 0.5 340
1500 =
0.1133 Ω, 𝑋𝑚 = 23.6 340
1500 = 5.35 Ω.
Scalar v/f control using VSI: Example (2) End
Thus the equivalent circuit looks like the following:
V1
I1
R1=0.1 Ω
jX1=j0.068
ΩjX2
’=j0.1133 Ω
R2’=0.17 Ω
((1-s)/s)R2’ =0.99ΩjXm= j5.35 Ω
41.12<0
Im
Thus 𝐼𝑚 =41.12∠00(0.17+0.99+𝑗0.1133 )
𝑗5.35=8.96∠ − 84.42°A = 0.87 − 𝑗8.92 A.
∴ 𝐼1 = 𝐼𝑚 + 41.12 = 0.87 − 𝑗8.92 + 41.12 = 42.93∠−11.99° A
∴ 𝑉1 = 𝐼1 0.1 + 𝑗0.68 + 𝑗5.35 ∗ 𝐼𝑚 = 42.93∠−11.99° ∗ 0.121∠34.22° + 𝑗5.35 ∗ 8.96∠ −84.42° = 5.19∠22.23 + 47.94∠5.58° = 52.93∠7.19° V.
∴ Power − factor = cos 11.99° + 7.19° = 0.94 lag.
Efficiency = 𝑃𝑜𝑢𝑡
𝑃𝑖𝑛=
5000
3𝑉1𝐼1 cos 𝜃=
5000
3∗52.93∗42.93∗0.94= 0.78.
Induction Motor Speed up to Rated Speed Using Slip Speed Control
Pulse Width
Modulated
(PWM) Inverter
V1,ref
w1,ref
Function
generator
SCIMMechanical
load
+
+Slip Speed
Controller
-
+wref
ω ω
w1,ref
Limiter
Vd
Scalar v/f control using VSI: High Speed Region (From Rated Speed Up To 1.5-2 Times Rated Speed)
Beyond rated speed we cannot increase 𝑉1 since this is the rated value. As the mechanical
structure (bearings, rotor ) is usually sturdy increasing beyond rated speed is not a problem.
Strategy of speed control:
Above base speed the maximum allowable value of s is at its rated value (𝑠𝑟𝑎𝑡𝑒𝑑 ) and the supply
frequency is increased. Thus the maximum allowable steady state value of torque in this region is
𝑇𝑜𝑢𝑡 = |𝑉1|
𝜔1
2
∙𝑠𝑟𝑎𝑡𝑒 𝑑 𝜔1
𝑅2′ =
|𝑉1|2𝑠𝑟𝑎𝑡𝑒𝑑
𝑅2′ ∙
1
𝜔1=
|𝑉1|2𝑠𝑟𝑎𝑡𝑒𝑑
𝑅2′ ∙
1−𝑠𝑟𝑎𝑡𝑒𝑑
𝜔∝
1
𝜔.
Thus torque developed by the motor is inversely proportional to the speed. Therefore the motor
works in constant power region.
This region continues as long as the slip speed does not exceed the maximum allowable slip
speed 𝜔2 determined by the breakdown or maximum torque capability of the machine.
Scalar v/f control using VSI: Very High Speed Region (Beyond 1.5 or 2 times Rated Speed)
This region starts when slip speed reaches 𝜔2 = 𝑠𝑟𝑎𝑡𝑒𝑑 𝜔1 (the maximum allowable slip speed).
In this speed zone the slip speed is limited to 𝜔2 , the maximum allowable slip speed or the
breakdown slip speed. Thus the maximum allowable steady state value of torque in this region
𝑇𝑜𝑢𝑡 = |𝑉1|
𝜔1
2
∙𝑠𝑟𝑎𝑡𝑒𝑑 𝜔1
𝑅2′ =
|𝑉1|2𝜔2
𝑅2′ ∙
1
(𝜔1)2
With small value of 𝑠, the actual speed of the motor is very close to synchronous speed.Thus
torque is inversely proportional to the square of speed, while power is inversely proportional to
speed.
Summary of speed control philosophy in the three regions of operation
• Below rated (base) speed (Constant torque region)
1. Change slip speed to get required torque for any speed. 2. Obtain frequency as sum of speed and slip speed. 3. Determine supply voltage by adding stator resistance drop to the component that is proportional to frequency.
• Above rated (base) speed up to 1.5-2 times rated speed (Constant power region)
1. Keep supply voltage constant at rated value. 2. Limit slip to its rated value and adjust frequency such that slip speed does not exceed the breakdown or maximum allowable level of slip speed.
• Very high speed region beyond 1.5-2 times rated speed
1. Keep supply voltage constant at rated value. 2. Adjust frequency by limiting slip speed to its breakdown or maximum allowable level.
Scalar v/f control using VSI: Example (3) A 1740 rpm, 3 phase, 60 Hz, 4 pole, 460 V, 3 hp Y connected induction motor with 𝜔2 =
1.8𝜔1𝑠𝑟𝑎𝑡𝑒𝑑 is operated from an inverter and is expected to work at high/ very high speed
regions. Draw T vs. Speed, slip speed vs. Speed, slip vs. Speed, power vs. Speed indicating the
high speed and very high speed region.
Solution:
Synchronous speed 𝑁1 =120𝑓
𝑃=
120∗60
4= 1800 rpm.
∴ 𝑠𝑟𝑎𝑡𝑒𝑑 =1800−1740
1800= 0.033
𝜔1,𝑟𝑎𝑡𝑒𝑑 =2𝜋𝑁1
60=
2𝜋∗1800
60= 188.5 𝑟𝑎𝑑/𝑠
Therefore, rated slip speed = 𝑠𝑟𝑎𝑡𝑒𝑑 𝜔1,𝑟𝑎𝑡𝑒𝑑 = 6.22 𝑟𝑎𝑑/𝑠
𝑇𝑟𝑎𝑡𝑒𝑑 =𝑃𝑜𝑢𝑡 ,𝑟𝑎𝑡𝑒𝑑
𝜔=
3×746
2𝜋×1740/60=
2238
182.21= 12.28 𝑁𝑚
Transition to very high speed region takes place when 𝜔2 = 1.8𝑠𝑟𝑎𝑡𝑒𝑑 𝜔1,𝑟𝑎𝑡𝑒𝑑 . Therefore,
𝜔2 = 1.8 ∗ 6.22 = 11.2 𝑟𝑎𝑑/𝑠. Corresponding synchronous speed 𝜔1 =𝜔2
𝑠𝑟𝑎𝑡𝑒𝑑= 339.4 𝑟𝑎𝑑/𝑠.
Motor speed corresponding to this synchronous speed in rpm is given by
𝑁 =𝜔1 1−𝑠𝑟𝑎𝑡𝑒𝑑
2𝜋
60
= 3134 rpm . Torque at this speed is 𝑇 =𝑃𝑜𝑢𝑡 ,𝑟𝑎𝑡𝑒𝑑
𝜔=
2238
3134∗ 2𝜋
60
=2238
328.19=
6.82 Nm.
Scalar v/f control using VSI: Example (3) Contd...
The following figure shows the variation of the different parameters in the different speed zones
in details.
Torque vs. speed
12.3
1740 3134
6.8
Speed (rpm)
T (Nm)
Slip speed vs. speed
slip speed
(rad/s)
1740 3134 Speed (rpm)
6.2
11.2
Power vs. speed
P (W)
1740 3134 Speed (rpm)
2238
Slip vs. speed
1
1740 3134 Speed (rpm)
0.033
Slip
Scalar v/f control using Current Source Inverter (CSI)
R1
jX1 jX2’
R2’
((1-s)/s)R2’
jXm
I2’
ImI1
Recall that, neglecting rotational losses, 𝑇𝑜𝑢𝑡 = 𝑃𝑜𝑢𝑡
𝜔=
𝑃𝑜𝑢𝑡
(1−𝑠)𝜔1=
𝑃𝑎𝑔
𝜔1= 𝐼2
′ 2 𝑅2′
𝑠𝜔1
Now 𝐼2′ =
𝑗𝑋𝑚
𝑅1 +𝑅2
′
𝑠+ 𝑗(𝑋1 + 𝑋2
′ + 𝑋𝑚)
𝐼1
∴ 𝑇𝑜𝑢𝑡 = 𝐼2′ 2 𝑅2
′
𝑠𝜔1=
𝑥𝑚2
𝑅1+𝑅2′
𝑠
2
+(𝑋1+𝑋2′ +𝑋𝑚 )2
𝑅2′
𝑠𝜔1 𝐼1
2 ≈1
𝑅2′ .
𝑠𝜔1𝐿𝑚2
1+(𝐿1+𝐿2′ +𝐿𝑚 )2 𝑠𝜔 1 2
𝑅2′ 2
𝐼1 2; assuming 𝑅1 ≪
𝑅2′
𝑠.
Thus torque can be controlled by varying slip speed 𝑠𝜔1 as determined from the slip speed controller. Then
the following relationship between 𝐼1 and 𝐼𝑚 can be used to determine 𝐼1 for a given slip speed.
To maintain flux at rated value at all speed up to base speed
𝐼𝑚 ≈
𝑅2′
𝑠
2
+ 𝑋1+𝑋2′
2
𝑅2′
𝑠
2
+ 𝑋1+𝑋2′ +𝑋𝑚
2
𝐼1 =
𝑅2′
𝑠𝜔 1
2
+ 𝐿1+𝐿2′
2
𝑅2′
𝑠𝜔 1
2
+ 𝐿1+𝐿2′ +𝐿𝑚
2
𝐼1 = constant ; assuming 𝑅1 ≪𝑅2
′
𝑠
Thus 𝐼1 =
𝑅2′
𝑠𝜔 1
2
+ 𝐿1+𝐿2′ +𝐿𝑚
2
𝑅2′
𝑠𝜔 1
2
+ 𝐿1+𝐿2′
2
𝐼𝑚
Scalar v/f control using Current Source Inverter (CSI) Contd...
From previous slide:
𝑇𝑜𝑢𝑡 =𝑥𝑚
2
𝑅1 +𝑅2
′
𝑠 2
+ (𝑋1 + 𝑋2′ + 𝑋𝑚)2
𝑅2′
𝑠𝜔1
𝐼1 2
To obtain the maximum torque, the above expression is differentiated with respect to slip and
equated to zero. This leads to
𝑇𝑚𝑎𝑥 =|𝐼1|2
𝑅1+ 𝑅12+ 𝑋1+𝑋2
′ +𝑋𝑚 2
∙1
2𝜔1 ≈
𝐼1 2
2𝜔1 𝑋1+𝑋2′ +𝑋𝑚
Slip at which maximum torque occurs
𝑠𝑇𝑚𝑎𝑥 =𝑅2
′
𝑅12 + 𝑋1 + 𝑋2
′ + 𝑋𝑚 2
≈𝑅2
′
𝜔1 𝐿1 + 𝐿2′ + 𝐿𝑚
Scalar v/f control using Current Source Inverter (CSI) End
Because of the presence of the extra term 𝑋𝑚 in the denominator of the 𝑠𝑇𝑚𝑎𝑥 expression the
𝑠𝑇𝑚𝑎𝑥 value is much lower compared to being driven by a VSI (see figure below). Consequently,
the region of stable operation between the s = 0 and s = 𝑠𝑇𝑚𝑎𝑥 becomes very steep. Thus the
current source inverter driven induction motors must be operated with the speed loop closed.
slip s = 0
s = 1
T
Tmax
VSI
TL
Tm1 Tm2 Tm3
slip s = 0
s = 1
T
Tmax CSI
TL
Tm1 Tm2 Tm3
Compare VSI and CSI sTmax :Example (4) A 450 V, 50 Hz, 1450 rpm, 25 kW, Y connected, 3 phase induction motor delivers rated torque
at all speed. The motor equivalent circuit parameters at rated frequency are 𝑅1 = 0.1 Ω, 𝑅2, =
0.17 Ω, 𝑋1 = 0.3 Ω, 𝑋2′ = 0.5 Ω, 𝑋𝑚 = 23.6 Ω. Compare the slip speed at which 𝑇𝑚𝑎𝑥 can be
obtained for i) a VSI and ii) a CSI.
Solution:
i) For a VSI slip speed at 𝑇𝑚𝑎𝑥 is given by
𝜔1𝑠𝑇𝑚𝑎𝑥 ≈𝑅2′
𝐿1+𝐿2′
=0.17∗2𝜋∗50
0.3+0.8 = 485.5 rad/s . For 50 Hz operation this would mean a
𝑠𝑇𝑚𝑎𝑥 =485.52
2𝜋∗50= 0.1545.
ii) For a CSI slip speed at 𝑇𝑚𝑎𝑥 is given by
𝜔1𝑠𝑇𝑚𝑎𝑥 ≈𝑅2′
𝐿1+𝐿2′ +𝐿𝑚
=0.17∗2𝜋∗50
0.3+0.8+23.6 = 2.16 rad/s . For 50 Hz operation this would mean a
𝑠𝑇𝑚𝑎𝑥 =2.16
2𝜋∗50= 0.0069.
Current Regulated Induction Motor Drive