1. Electro Mechanical System
Active Power Flow
Let us see how electrical energy is converted into mechanical energy.
Active power Pe flows from the line into 3-phase stator.
Due to stator copper losses Pjs is dissipated as heat in the windings.
Another portion Pf is dissipated as heat in the stator, iron losses.
Remaining active power Pr is carried across the air gap and transferred to the rotor by electromagnetic induction.
Pjr is dissipated as heat due to I2 R losses in the rotor.
Pm is final mechanical power available.
Pv is another small portion representing windage & bearing friction losses
We finally get PL available at the shaft to drive the load.
Important properties of induction motor: 1. Efficiency 2. Power 3. Torque
Lecture 22
Electro Mechanical System

2. Electro Mechanical System
Efficiency
Efficiency is the ratio of output power to the input power: 
Efficiency (η) = PL / Pe
I2R losses in the rotor
Rotor I2 R losses or Pjr are related to the rotor input power Pr as:
Pjr = sPr
Where:
Pjr = Rotor I2R losses [W]
s = slip
Pr = Power Transmitted to the rotor [W]
As the slip increases, the rotor I2R losses consumes a larger and larger portion of power Pr across the air gap to the rotor.
A rotor turning at half synchronous speed (s = 0.5).
50% active power received, dissipates in the form of heat
When rotor is locked (s =1), all the power transmitted to the rotor is dissipated in the form of heat.
Lecture 22
Electro Mechanical System

3. Electro Mechanical System
Mechanical Power
Mechanical power Pm developed by the motor is equal to power transmitted to the rotor minus I2R losses:
Pm = Pr – Pjr
Since:
Pjr = sPr So
Pm = Pr – sPr
Pm = (1 – s)Pr
Actual mechanical power available is slightly less than Pm due to windage and friction losses.
In most calculations we can neglect this.
Lecture 22
Electro Mechanical System

4. Electro Mechanical System
Motor Torque
Mechanical power output of a motor depends upon its rotational speed and torque it develops. The power of a motor is given by:
Pm = Pr – Pjr =(nTm)/9.55
Where:
P = Mechanical Power (Watts)
n = speed of rotation (rpm)
Tm = Torque (N.m)
9.55 = Constant to take care of the units (exact value 30/)
Considering the above equation the mechanical torque Tm developed by induction motor at any speed is given by:
Pr = Power transmitted to the rotor(Watts)
ns = synchronous speed (rpm)
9.55 = Multiplier(exact value 60/2)
T = Torque (N.m)
Lecture 22
Electro Mechanical System

5. Electro Mechanical System
Motor Torque
Actual Torque TL available at the shaft is slightly less than Tm due to windage and friction losses.
We neglect this small difference.
Torque equation shows that torque(Tm) is directly proportional to the active power transmitted to the rotor.
To develop a high locked rotor torque, the rotor must absorb a large amount of active power.
This absorbed power is dissipated as heat, so the motor temperature rises very rapidly.
Lecture 22
Electro Mechanical System

6. Torque versus Speed curve
Torque developed by a motor depends upon its speed. Relationship between the two can be expressed by the above curve
3-Phase induction motor has full load torque T & starting torque 1.5T
Max. torque is called breakdown torque of the motor is 2.5 T. If load is increased beyond, the motor will stall and come to a rapid stop.
At full load the motor runs at a speed n.
Lecture 22
Electro Mechanical System

7. Torque versus Speed curve
If the mechanical load increases, the speed will drop until the motor torque is equal to load torque. when the two torques are in balance, the motor will run at a constant but slightly lower speed.
Small motors (10 kW) will develop breakdown torque at a speed nd at about 80% of synchronous speed ns.
Big motors (1000 kW and more ) attain their breakdown torque at about 98% of synchronous speed ns.
Lecture 22
Electro Mechanical System

8. Effect of Rotor resistance
The rotor resistance of a squirrel cage rotor is essentially constant, except that it increases with temperature. The resistance increases with increase in load because the temperature rises.
In designing the squirrel cage rotor, the rotor resistance can be selected by using copper, Aluminum or other metals in the rotor bars and end rings.
The only characteristic which is unchanged is the breakdown torque.
Lecture 22
Electro Mechanical System

9. Effect of Rotor resistance
Following diagram shows the torque speed curve of a 10kW (13.4hp), 50Hz, 380V induction motor having a synchronous speed of 1000rpm and a full load torque of 100N.m. Full load current is 20A and locked rotor current is 100A. The rotor has an arbitrary resistance R.
Lecture 22
Electro Mechanical System

10. Effect of Rotor resistance
Increase the rotor resistance by a factor of 2.5 by using higher resistive material such as bronze for rotor bars and end rings.
The starting torque doubles in the new torque speed curve
The locked rotor current decreases from 100A to 90A.
Motor develops its breakdown torque at a speed nd of 500 rpm. Compared to original nd of 800 rpm.
Lecture 22
Electro Mechanical System

11. Effect of Rotor resistance
If we double the resistance so that it becomes 5R, the locked rotor torque attains a maximum value of 250 Nm for a corresponding current of 70A
further increase in resistance will drop the locked rotor current and torque. If we increase it to 25R, the locked rotor current drops to 20A
The motor will develop the same starting torque 100Nm as it did when the locked rotor current was 100A
Lecture 22
Electro Mechanical System

12. Effect of Rotor resistance
Summary:
A high rotor resistance is desirable because it produces a high starting torque and a relatively low starting current.
Unfortunately it produces a rapid fall in speed with increasing load.
Slip at rated torque is high, the motor I2 R losses are high.
Efficiency is low and motor tends to overheat.
Under running conditions it is preferable to have low resistance.
The speed decreases very less with increasing load.
Slip at rated torque is small.
The efficiency is high and the motor tends to run cool.
Lecture 22
Electro Mechanical System