Performance Characteristics. 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
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- power factor
- stator current
- starting torque
- maximum torque (pull-out), etc
At high slip
At low slip
Differentiate eqn. dT/ds, and equate to zero
If R1 small
Increase R2, increase slip max, increase staring torque
If R1 small
Stator current vs. speed
Power factor vs. speed
Efficiency vs. speed
*To get Max. efficiency, s must be very low
0 < s < 1
0 < s
s > 0
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,
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?
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 short-circuited. 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 developed torque Tm = load torque TL
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
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 %)
Sen 282 (pb 5.6)
Synchronous speed change with changes number of poles (change the stator winding/coil connection)
Discrete step change in speed/ expensiveSpeed 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
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 (amplitude), frequency fixed
E.g. using 3-phase autotransformer (variac) or solid state controller
Slip becomes high as voltage reduced – low efficiency
Fan (TL) load
Solid State Voltage Control
Closed Loop Operation Voltage Control
Open Loop Control Scheme
Closed Loop Control Scheme
Control both V and freq, f
3-phase pwm Inverter
Rectifier and FilterTypical IM Drive System - Variable voltage, variable frequency
SESI 2007/2008 eg 5.4 pg 241