UTILIZATION OF ELECTRICAL ENERGY
CHAPTER 3: Control of Electric Drives
L-3-5
1
AC Drive Control
3-Phase Induction Motor
• Induction Motors are simple in design, rugged in construction and reliable in
service as they have no commutation.
• Besides they have low initial cost, easy operation & simple maintenance, high
efficiency and simple control gear for starting and speed control.
• The energy is transferred from primary(stator) to secondary(rotor) entirely by
induction. Therefore induction motor is essentially a transformer(short circuited
secondary, rotating transformer).
• At standstill, induction motor is actually a static transformer having its
secondary winding short circuited.
Contd.....
• In the transformer, the load on secondary is electrical and in an induction motor the load is mechanical
which can be replaced by an equivalent electrical load of resistance RL.
Since, I2 = =
/
R2 = resistance per phase of rotor & k is the turn ratio of secondary to primary
If R2’ is the effective rotor resistance referred to stator, then the combination with RL is
R2’/s = R2’ + R2’ − 1 = R2’+(RL)’
Contd.....
The simplified equivalent circuit of 3-ph induction motor is shown below.
The rotor current referred to stator is,
I2’ =
/
Losses = s*Air gap Power
Or, 3*(I2’)2*R2’ = s* *T
3 (I2’)2 R2’
Or, T = ; ns = rps
∗
Also,
/
T=
Contd....
3 V2 R2’/s
T= Nm
∗ [ ]
By differentiating the above equation w.r.t s and equating to zero we get the slip
corresponding to maximum torque as
s=
Then the maximum torque will be,
Tmax =
[ ± ]
+ for motoring
- generating
Example
Example
Solution: 400V, 4 pole, 50 hz, 3-phase, Y
N1 = 2N2 ; N2/N1 = ½
r1 = 0.64Ω x1 = 1.1Ω r2 = 0.08Ω x2 = 0.12Ω
TL N2
TL1 = 40 Nm at N1 = 1440 rpm
N2 = 1300 rpm, TL2 = ?
; TL2 = 32.6 rpm
b. r2’ = r2/k2 = 0.08/(1/2)2 = 0.32Ω
x2’ = x2/k2 = 0.12/(1/2)2 = 0.48 Ω
Contd....
We know,
𝑹𝒐𝒕𝒐𝒓 𝒄𝒐𝒑𝒑𝒆𝒓 𝒍𝒐𝒔𝒔/𝒔
Torque = [OK since T = ]
𝟐𝝅𝒏𝒔
3 (I2’)2 r2’
Or, 32.6 = ∗
Or, I2’ = 26.635 A
Actual rotor current = I2’/k = 26.635*2 = 53.27 A
Also,
I2’ =
/
Or, V = 91.395 V
Stator voltage (L-L) = *91.935 = 158.3 V
Ns = 120f/P = 1500 rpm Nr = 1300 rpm
s = 0.133 = (Ns-N)/Ns
Reduced Voltage Starting(Soft Start)
The starting line current at full voltage of an induction motor can be about 6
times the rated full load current (why?).
Severe Voltage Drop may occur in the network supplying the induction motor.
Reduced voltage starting can be used at the starting as illustrated in given
circuit.
Contd.....
• By proper control of firing angle, a low output voltage can be supplied to the
induction motor.
• When the motor reaches full speed, the regulator can be short circuited by
mechanical contractor so that the motor operates at rated voltage.
Slip Power Recovery Scheme
Background: Pm = Pg-Pr ; Pr = sPg
When an induction motor is in high slip, i.e. the rotor is running much slower than
the stator field is rotating at, then a large amount of rotor current flows. Slip
recovery uses this rotor current to convert this otherwise wasted energy back to the
power supply system
The slip power from the rotor circuit can be recovered and fed back to the AC
source so as to utilize it outside the motor. Thus, the overall efficiency of the drive
system can be increased.
Since the frequency of rotor currents is slip frequency, this method is known as slip
power recovery scheme.
(not possible is squirrel cage rotor as the rotors bars are short circuited at the ends)
Static Scherbius Drive
Contd....
For achieving both sub synchronous & super synchronous speed control, both
the converters must be fully controlled.
The three phase transformer is meant to bring the rotor circuit at a value
suitable to be fed back into the supply.
The major drawback is the poor power particularly at low speeds, as the
bridges draw reactive power from the mains.
Finds application in large power fan and pump drives which need speed control
in a narrow range only.
The cost of converter is appreciable and a slip frequency gating circuit is
required.
Detailed Note: JB Gupta 3.13.7
Static Kramer Drive
Contd....
The rotor circuit feeds the armature of separately excited dc motor(which is
mechanically coupled to the IM) rectified by diode bridge.
Speed control is achieved by varying the field current of dc motor.
For achieving large speed range, diode bridges are replaced with thyristor
bridge.
The absence of inverter causes it to draw less reactive power, hence pf is
improved as compared to Static Scherbius drive and introduces low harmonic
content.
Maintenance problems due to commutator and brushes and has a drawback of
large inertia.
End of L-3-5
