Induction Motor characteristics

AIM : To study about the characteristics of Induction Motor and compare induction motor to a mechanical clutch.

CHALLENGES :

1. How induction motor operation is similar to the clutch?

First we will see the working of clutch and induction motor separately.

Mechanical Clutch –

Clutch works on the principle of friction because when the two rough surface clutch plates are comes in contact with each other then they become rotating as a single unit. The clutch is located between the engine and the gearbox, mounted directly to the face of the engine's flywheel.

The clutch connects and disconnect two rotating shafts (drive shafts or line shafts), one shaft is typically attached to an engine (the driving member) while the output shaft (the driven member) provides output power for work.

The default state of the clutch is engaged, i.e. the connection between engine and gearbox is always on unless the driver presses the pedal and disengages it. If the engine is running with clutch engaged and the transmission in neutral, the engine spins the input shaft of the transmission, but no power is transmitted to the wheels. Although the gearbox does not stop rotating during a gear change, there is no torque transmitted through it.

The output shaft of the gearbox is permanently connected to the final drive, then the wheels, and so both always rotate together. With the clutch disengaged, the gearbox input shaft is free to change its speed as the internal ratio is changed. Any resulting difference in speed between the engine and gearbox is evened out as the clutch slips slightly during re-engagement.

Induction Motor –

The stator and rotor are two essential parts of the motor. The stator is the stationary part, and it carries the overlapping windings while the rotor carries the main or field winding. The windings of the stator are equally displaced from each other by an angle of 120°.

The induction motor is the single excited motor, i.e., the supply is applied only to the one part, i.e. stator. The term excitation means the process of inducing the magnetic field on the parts of the motor. When the three phase supply is given to the stator, the rotating magnetic field produced on it. The torque induced on the conductor is known as electromagnetic torque.

The direction of electromagnetic torque and rotating magnetic field is same. Thus, the rotor starts rotating in the same direction as that of the rotating magnetic field.

Comparing Mechanical Clutch to Induction Motor :

As mentioned above, the driving shaft and driven shaft attached to the clutch are rotating at different speeds which cause a slip and clutch attempts to lock them together or release them. The slip that exist between the driver and the driven shaft is given by,

`ω_2=ω_1 (1-s)`

Where,

s – Slip;

`omega_1` – Speed of driving shaft;

`omega_2` – Speed of driven shaft.

The operation of clutch is similar to that of an induction motor in terms of the slip. In an induction motor the speed of the rotor is always less than the rotating magnetic field or synchronous speed. The rotor tries to the run at the speed of the rotor, but it always slips away. Thus, the motor never runs at the speed of the rotating magnetic field, and this is the reason because of which the induction motor is also known as the asynchronous motor. If the speed of the rotor is equal to the synchronous speed, no relative motion occurs between the rotating magnetic field of the stator and the conductors of the rotor. Thus the EMF is not induced on the conductor, and zero current develops on it. Without current, the torque is also not produced. Because of the above mention reasons the rotor never rotates at the synchronous speed. The speed of the rotor is always less than the speed of the rotating magnetic field. This is why the slip occurs and it is given as,

`s= (N_s-N_r)/N_s`

Where,

s – Slip;

N_s – Synchronous Speed;

N_r – Speed of the Rotor;

The above equation can be rearranged as follows,

`N_r=N_s (1-s)`

This states that the operation of induction motor is similar to that of clutch.

2. Calculate the starting time of a drive with the following parameters:

J = 10 kg-m²,

T_m = 15 + 0.5`omega_m`,

T_l = 5 + 0.6`omega_m`,

Solution :

Here,

J – Moment of inertia of Motor and Load,

T_m – Motor Torque,

T_l­ – Load Torque,

The steady state stability equation for an induction motor is given by,

`T_m – T_l=J×(dω)/dt` ----- (1)

Where,

d`omega` = Change in angular velocity;

dt = Change in time.

At steady state, the motor torque is equal to the load torque,

`therefore``T_l=T_m`

`15+0.5ω=5+0.6ω`

`0.1ω = 10`

`therefore``omega` = 100 rad/s

The steady state speed for the induction motor is 100 rad/s

For calculating the starting time, the values of T_m, T_l, J and `omega` values substituted in equation (1),

`(15+0.5ω)-(5+0.6ω)=10×(dω)/dt`

`(dω)/dt=(10-0.1ω)/10`

`(dω)/dt=1-0.01ω`

`dt/(dω)=1/((1-0.01ω) )`

`dt=1/((1-0.01ω) )×dω`

Assuming 5% of slip the equilibrium speed will decrease from 100 rad/s to 95 rad/s

To get the time, we will integrate the above equation within the intervals,

`omega_1` = 0 rad/s

`omega_2` = 95 rad/s

`therefore``intdt= int_0^95 1/((1-0.01ω) )×dω`

`therefore``t=(1/0.01)×ln⁡|1-0.01ω|+C`

Solving for integration constant C by giving initial condition values,

Initial conditions, when t =0, `omega` = 0;

`therefore``0=(1/0.01)×ln⁡|1-0.01(0)|+C`

`therefore` C = 0

Now we can apply the interval values to get the time value,

`therefore``t=[(1/0.01)×ln⁡|1-0.01ω| ]`

`therefore``t=(2.99573×100)-0`

`therefore``t=299.5` seconds

So the starting time of induction motor (t) = 299.5 seconds

3. A drive has following equations for motor and load torques:

T_m = (1 + 2ω_m)

T_l = 3 √ω_m

Obtain the equilibrium points and determine the stability.

Solution :

The stability of the motor is the capacity of the motor to drive the load. If the motor has enough power to drive the load then it is called as the stable motor.

The steady state stability equation for an induction motor is given by,

`T_m – T_l=J×(dω)/dt`

Here the motor torque is given as,

`T_m=1+2ω_m`

And the load torque is given as,

`T_l=3sqrt(ω_m )`

At steady state, the motor torque is equal to the load torque,

`therefore``T_l=T_m`

`1+2ω_m=3sqrt(ω_m )`

`2ω_m-3sqrt(ω_m )+1=0`

By solving the above equation, we get the equilibrium points as,

`sqrt(ω_m )` = 0.5 or 1; and `omega_m` = 0.25 or 1

The stability condition for induction motor is given by the difference between changes in Load torque with respect to speed and changes in Motor torque with respect to speed. It should be greater than 0 for motor drive to be stable.

`therefore``(dT_l)/(dω)-(dT_m)/(dω)>0`

Now differentiating,

`(dT_l)/(dω)=(d(3sqrt(ω_m )))/(dω)=3/(2sqrt(ω_m ))`

And `(dT_m)/(dω)=(d(1+2ω_m ))/(dω)=2`

`therefore``3/(2sqrt(ω_m ))-2>0`

Checking the above conditions for values of `sqrt(ω_m )`;

For `sqrt(ω_m )=0.5`,

`LHS=3/(2(0.5)) -2`

`LHS=1>0`, Therefore the condition is satisfied.

For `sqrt(ω_m )=1`,

`LHS=3/(2(1)) -2`

`LHS=-0.5<0`, Therefore the condition is not satisfied.

Thus the stability is achieved at point `sqrt(ω_m )=0.5` or where speed `omega_m` = 0.25 rad/s.